Give the equations of two lines passing through (4, –2). How many more such lines are there, and why? -Maths 9th

Give the equations of two lines passing through (4, –2). How many more such lines are there, and why? -Maths 9th
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A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Description : A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Last Answer : Clearly, line a is parallel to X-axis at a distance of 1 unit in positive direction of Y-axis, therefore its equation is y = 1. Also, if we draw the graph of line 2x + 3y = 6, then its graph should intersect X - axis at (3,0 ... Base Height = 1/2 BC AC = 1/2 1 3 / 2 = 3 / 4 sq unit.

A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Description : A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

Last Answer : Clearly, line a is parallel to X-axis at a distance of 1 unit in positive direction of Y-axis, therefore its equation is y = 1. Also, if we draw the graph of line 2x + 3y = 6, then its graph should intersect X - axis at (3,0 ... Base Height = 1/2 BC AC = 1/2 1 3 / 2 = 3 / 4 sq unit.

What is the angle between the lines whose equations are: 3x + y – 7 = 0 and x + 2y + 9 = 0. -Maths 9th

Description : What is the angle between the lines whose equations are: 3x + y – 7 = 0 and x + 2y + 9 = 0. -Maths 9th

Last Answer : (c) (8, 6)Let AB be the given line 4x + 3y = 25 Let O′(a, b) be the image of O in the given line AB. Let O O′ cut AB in point P. Also OP ⊥ AB and P is the mid-point of OO′. ∴ Co-ordinates of P are \(\bigg( ... 4 imes6}{3}\) = 8∴ The image of the point O(0, 0) in the line 4x + 3y - 25 = 0 is (8, 6).

ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D. -Maths 9th

Description : ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D. -Maths 9th

Last Answer : According to question p rove that ∠CBD +∠CDB = 1/2 ∠BAD.

ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D. -Maths 9th

Description : ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D. -Maths 9th

Last Answer : According to question p rove that ∠CBD +∠CDB = 1/2 ∠BAD.

Write three possible linear equations which can pass through point (3, –2). -Maths 9th

Description : Write three possible linear equations which can pass through point (3, –2). -Maths 9th

Last Answer : Solution :-

If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Description : If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Last Answer : According to question the radius of the circle passing through the points A, B and C .

If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Description : If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Last Answer : According to question the radius of the circle passing through the points A, B and C .

Write the equation of a line parallel to y-axis and passing through the point (–4, –5). -Maths 9th

Description : Write the equation of a line parallel to y-axis and passing through the point (–4, –5). -Maths 9th

Last Answer : Solution :- x= -4

Write the equation of a line parallel to x-axis and passing through the point (–3, –4). -Maths 9th

Description : Write the equation of a line parallel to x-axis and passing through the point (–3, –4). -Maths 9th

Last Answer : Solution :- y = -4

Find the centre of a circle passing through the points (6, –6), (3, –7) and (3, 3). -Maths 9th

Description : Find the centre of a circle passing through the points (6, –6), (3, –7) and (3, 3). -Maths 9th

Last Answer : For three points to be collinear, the area of the triangle formed by the three points should be zero. ∴ Area of D formed by the given three points= \(rac{1}{2}\) [a((c + a) - (a + b)) + b((a + b) - (b + c)) + c((b ... ba - bc + cb - ca] = 0.Hence (a, b + c), (b, c + a) and (c, a + b) are collinear.

Find the value of a, if the line passing through (–5, –8) and (3, 0) is parallel to the line passing through (6, 3) and (4, a). -Maths 9th

Description : Find the value of a, if the line passing through (–5, –8) and (3, 0) is parallel to the line passing through (6, 3) and (4, a). -Maths 9th

Last Answer : Let C(x, y) be the centre of the circle passing through the points P(6, -6), Q(3, -7) and R(3, 3) Then, PC = QC = RC(Being radius of the same circle) PC2 = QC2 ⇒ (x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2 ⇒ x2 - ... i), we get 3\(x\) + (-2) - 7 = 0 ⇒ 3\(x\) = 9 ⇒ \(x\) = 3 ∴ The centre is (3, -2).

Find the equation of the straight line passing through the point (4, 5) and perpendicular to 3x – 2y + 5 = 0. -Maths 9th

Description : Find the equation of the straight line passing through the point (4, 5) and perpendicular to 3x – 2y + 5 = 0. -Maths 9th

Last Answer : There are two ways to prove it. 1st way: Area of triangle formed by the given points = 0 if they are collinear.∴ \(rac{1}{2}\) [\(x\)(2 - (y + 1) ) + 1((y + 1) - 1) + 0(1 - 2)] = 0⇒ \(rac{1}{2}\) [2\(x\) - \(x\)y - \( ... y - \(x\)y - 1 + \(x\) ⇒ x + y = \(x\)y ⇒ \(rac{1}{x}\) + \(rac{1}{y}\) = 1.

What is the equation of the straight line passing through the point (4, 3) and making intercepts on the co-ordinates axes whose sum is –1 ? -Maths 9th

Description : What is the equation of the straight line passing through the point (4, 3) and making intercepts on the co-ordinates axes whose sum is –1 ? -Maths 9th

Last Answer : Diagonals of a rhombus bisect each other at right angles ⇒ Co-ordinates of mid-points of AC and BD are equal∴ 0 = \(\bigg(rac{4+(-2)}{2},rac{-5+(-1)}{2}\bigg)\) = (1, -3)Slope of BD = \(rac{-5+1}{4+2}\) = \(rac{-4}{6}\) ... (rac{3}{2}\) isy + 3 = \(rac{3}{2}\) (x - 1)⇒ 2y + 6 = 3x - 3 ⇒ 2y = 3x - 9.

What is the equation of the line passing through (2, –3) and parallel to the y-axis ? -Maths 9th

Description : What is the equation of the line passing through (2, –3) and parallel to the y-axis ? -Maths 9th

Last Answer : (c) x = 2 Slope of y-axis = tan 90º (∵ y-axis ⊥ x-axis) ∴ Equation of line passing through (2, –3) parallel to y-axis is (y + 3) = tan 90º (x – 2) ⇒ (y + 3) = ∞ (x – 2) ⇒ (x – 2) = \(rac{1}{\infty}\) (y + 3) = 0 ⇒ x = 2.

A transversal intersects two lines in such a way that the two interior angle on the same side of transversal are equal.Will the two lines always be parallel? -Maths 9th

Description : A transversal intersects two lines in such a way that the two interior angle on the same side of transversal are equal.Will the two lines always be parallel? -Maths 9th

Last Answer : Solution :- The two lines will not be always parallel as the sum of the two equal angles will not always be 180°. Lines will be parallel when each of the equal angles is equal to 90°.

l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF

l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF

l,m and n are three parallel lines intersected by transversal p and q such that l,m and n cut-off equal intersepts AB and BC on p (Fig.8.55). Show that l,m and n cut - off equal intercepts DE and EF on q also. -Maths 9th

Description : l,m and n are three parallel lines intersected by transversal p and q such that l,m and n cut-off equal intersepts AB and BC on p (Fig.8.55). Show that l,m and n cut - off equal intercepts DE and EF on q also. -Maths 9th

Last Answer : Given:l∥m∥n l,m and n cut off equal intercepts AB and BC on p So,AB=BC To prove:l,m and n cut off equal intercepts DE and EF on q i.e.,DE=EF Proof:In △ACF, B is the mid-point of ... a triangle, parallel to another side, bisects the third side. Since E is the mid-point of DF DE=EF Hence proved.

Linear Equations in Two Variables Class 9th Formulas -Maths 9th

Description : Linear Equations in Two Variables Class 9th Formulas -Maths 9th

Last Answer : Rational Numbers : Rational Numbers are numbers , which can be expressed in the form p/q, q≠0; where p and q are integers . The collection of all rationals are represented by Q. ∴ Q = {p/q ; p,q ... number of rational numbers x2 , x1, x3,... between any two rational numbers a and b so that a

MCQ Questions for Class 9 Maths Chapter 4 Linear Equations in Two Variables with answers -Maths 9th

Description : MCQ Questions for Class 9 Maths Chapter 4 Linear Equations in Two Variables with answers -Maths 9th

Last Answer : 1) The linear equation 3x-11y=10 has: a. Unique solution b. Two solutions c. Infinitely many solutions d.No solutions c Explanation: 3x-11y=10 y=(3x-10)/11 Now for infinite values of x, y will ... c=0, always lie in the: a. First quadrant b. Second quadrant c. Third quadrant d. Fourth quadrant a

If two equations x^2 + a^2 = 1 – 2ax and x^2 + b^2 = 1 – 2bx have only one common root, then -Maths 9th

Description : If two equations x^2 + a^2 = 1 – 2ax and x^2 + b^2 = 1 – 2bx have only one common root, then -Maths 9th

Last Answer : x2+b2=1−2bx ...... (i) x2+b2+2bx=1 (x+b)2=1 x+b= 1 ∴x=1−b,−1−b x2+a2=1−2ax ...... (ii) x2+a2+2ax=1 (x+a)2=1 ∴x=1−a,−1−a It is given that the equations have only one root in ... or −1−a=−b−1 we get a=b but then both roots will be common, which is not possible. Hence, options A,B and C are correct.

Cbqs (case base study ) of chapter 4 Linear Equations in Two Variables of maths class 9th -Maths 9th

Description : Cbqs (case base study ) of chapter 4 Linear Equations in Two Variables of maths class 9th -Maths 9th

Last Answer : answer:

Check whetherx =2 and y = 1 is a solution of the following equations or not. -Maths 9th

Description : Check whetherx =2 and y = 1 is a solution of the following equations or not. -Maths 9th

Last Answer : Given, x = 2 and y = 1 (i) Given, linear equation is 2x + 5y = 9. On putting x = 2 and y= 1 in LHS, we get LHS = 2x + 5y =2(2) + 5(1) = 4 + 5 = 9 = RHS So, x = 2 and y=1 is a solution of given ... + 3 (1) = 5 + 3 = 8 ≠ 14 ⇒ LHS ≠ RHS So, x = 2 and y = 1 is not a solution of given equation.

In the following equations , find which of the variables x, y, z etc. represent rational numbers and which represent irrational numbers -Maths 9th

Description : In the following equations , find which of the variables x, y, z etc. represent rational numbers and which represent irrational numbers -Maths 9th

Last Answer : Following are the rational numbers which represent irrational numbers .

In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Description : In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Last Answer : (i) If we substitute x = 4, we get L.H.S = x + 4 = 4 + 4 = 8 and R.H.S = 2x = 2 x 4 = 8 ∴ L.H.S = R.H.S. Hence, 4 is a solution of x + 4 = 2x. (ii) If we substitute y = 3, we get L.H.S = y - 7 = 3 - ... S = 2u + 7 = 2(5) + 7 = 10 + 7 = 17 ∴ L.H.S = R.H.S. Hence , 5 is a solution of 3u + 2 = 2u + 7

In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Description : In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Last Answer : (i) f we substitute x = √2, we get L.H.S. = 2x - 3 = 2(√2) - 3 = 2√2 - 3 and R.H.S = x / 2 - 2 = √2 / 2 - 2 ∴ L.H.S. ≠ R.H.S . Hence, √2 is not a solution of 2x - 3 = x / 2 - 2 (ii) If we substitute ... = 7 ∴ L.H.S. ≠ R.H.S . Hence , -1 is not a solution of 24 - 3 (u - 2) = u + 8

How many linear equations in x and y can be satisfied by x = 1 and y = 2 ? -Maths 9th

Description : How many linear equations in x and y can be satisfied by x = 1 and y = 2 ? -Maths 9th

Last Answer : (c) Let the linear equation be ax + by + c = 0. On putting x = 1 and y = 2, in above equation we get =s a + 2b + c = 0, where a, b and c, are real number Here, different values of a, b and ... a + 2b + c = 0. Hence, infinitely many linear equations in x and yean be satisfied by x = 1 and y = 2.

Draw the graphs of linear equations y = x and y = – x on the same Cartesian plane. -Maths 9th

Description : Draw the graphs of linear equations y = x and y = – x on the same Cartesian plane. -Maths 9th

Last Answer : The given equation is y = x. To draw the graph of this equations, we need atleast two points lying on the given line. For x = 1, y = 1, therefore (1,1) satisfies the linear equation y = x. For x = 4, y = 4, ... of y = - x. We observe that, the line y = x and y = - x intersect at the point 0(0, 0).

Check whetherx =2 and y = 1 is a solution of the following equations or not. -Maths 9th

Description : Check whetherx =2 and y = 1 is a solution of the following equations or not. -Maths 9th

Last Answer : Given, x = 2 and y = 1 (i) Given, linear equation is 2x + 5y = 9. On putting x = 2 and y= 1 in LHS, we get LHS = 2x + 5y =2(2) + 5(1) = 4 + 5 = 9 = RHS So, x = 2 and y=1 is a solution of given ... + 3 (1) = 5 + 3 = 8 ≠ 14 ⇒ LHS ≠ RHS So, x = 2 and y = 1 is not a solution of given equation.

In the following equations , find which of the variables x, y, z etc. represent rational numbers and which represent irrational numbers -Maths 9th

Description : In the following equations , find which of the variables x, y, z etc. represent rational numbers and which represent irrational numbers -Maths 9th

Last Answer : Following are the rational numbers which represent irrational numbers .

In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Description : In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Last Answer : (i) If we substitute x = 4, we get L.H.S = x + 4 = 4 + 4 = 8 and R.H.S = 2x = 2 x 4 = 8 ∴ L.H.S = R.H.S. Hence, 4 is a solution of x + 4 = 2x. (ii) If we substitute y = 3, we get L.H.S = y - 7 = 3 - ... S = 2u + 7 = 2(5) + 7 = 10 + 7 = 17 ∴ L.H.S = R.H.S. Hence , 5 is a solution of 3u + 2 = 2u + 7

In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Description : In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Last Answer : (i) f we substitute x = √2, we get L.H.S. = 2x - 3 = 2(√2) - 3 = 2√2 - 3 and R.H.S = x / 2 - 2 = √2 / 2 - 2 ∴ L.H.S. ≠ R.H.S . Hence, √2 is not a solution of 2x - 3 = x / 2 - 2 (ii) If we substitute ... = 7 ∴ L.H.S. ≠ R.H.S . Hence , -1 is not a solution of 24 - 3 (u - 2) = u + 8

How many linear equations in x and y can be satisfied by x = 1 and y = 2 ? -Maths 9th

Description : How many linear equations in x and y can be satisfied by x = 1 and y = 2 ? -Maths 9th

Last Answer : (c) Let the linear equation be ax + by + c = 0. On putting x = 1 and y = 2, in above equation we get =s a + 2b + c = 0, where a, b and c, are real number Here, different values of a, b and ... a + 2b + c = 0. Hence, infinitely many linear equations in x and yean be satisfied by x = 1 and y = 2.

Draw the graphs of linear equations y = x and y = – x on the same Cartesian plane. -Maths 9th

Description : Draw the graphs of linear equations y = x and y = – x on the same Cartesian plane. -Maths 9th

Last Answer : The given equation is y = x. To draw the graph of this equations, we need atleast two points lying on the given line. For x = 1, y = 1, therefore (1,1) satisfies the linear equation y = x. For x = 4, y = 4, ... of y = - x. We observe that, the line y = x and y = - x intersect at the point 0(0, 0).

What is distance between the graphs of the equations y = -1 and y = 3? -Maths 9th

Description : What is distance between the graphs of the equations y = -1 and y = 3? -Maths 9th

Last Answer : Solution :- 4 units.

How many linear equations satisfy x = 2 and y = -3? -Maths 9th

Description : How many linear equations satisfy x = 2 and y = -3? -Maths 9th

Last Answer : Solution :- Infinitely many equations.

Draw the graph of the following equations (1) Y =3x+2 (b) Y=x -Maths 9th

Description : Draw the graph of the following equations (1) Y =3x+2 (b) Y=x -Maths 9th

Last Answer : This answer was deleted by our moderators...

Solve the following equations for x and y. log100 |x+y| = 1/2, -Maths 9th

Description : Solve the following equations for x and y. log100 |x+y| = 1/2, -Maths 9th

Last Answer : (b) \(\bigg(rac{10}{3},rac{20}{3}\bigg)\). (+ 10, 20) log100 |x+y| = \(rac{1}{2}\) ⇒ |x + y| = 100\(^{rac{1}{2}}\)⇒ |x + y| = 10 as (-10 is inadmissible) ...(i) log10y - log10| x | = log1004⇒ log10 ... x < 0, then x = 10.∴ If x = \(rac{10}{3}\), then y = \(rac{20}{3}\) and if x = 10, y = 20.

Examine the nature of the roots of the equations: (i) 2x^2 + 2x + 3 = 0 -Maths 9th

Description : Examine the nature of the roots of the equations: (i) 2x^2 + 2x + 3 = 0 -Maths 9th

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Of the following quadratic equations, which is the one whose roots are 2 and – 15 ? -Maths 9th

Description : Of the following quadratic equations, which is the one whose roots are 2 and – 15 ? -Maths 9th

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Find the values of k for which the equations x^2 – kx – 21 = 0 and x^2 – 3kx + 35 = 0 will have a common root? -Maths 9th

Description : Find the values of k for which the equations x^2 – kx – 21 = 0 and x^2 – 3kx + 35 = 0 will have a common root? -Maths 9th

Last Answer : Let the common root be α ⟹α2−kα−21=0......(1) α2−3kα+35=0........(2) (1)−(2)⟹2kα=56⟹α=k28​Substituting α=k28​ in (1) (k28​)2−28−21=0 ⟹k=±4

Solve these following equations: (i) 3x + 3 = 15 (ii) 2y + 7 =19 -Maths 9th

Description : Solve these following equations: (i) 3x + 3 = 15 (ii) 2y + 7 =19 -Maths 9th

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The probability of student A passing examination is 3/7 and of student B passing is 5/7 Assuming the two events “A passes”, -Maths 9th

Description : The probability of student A passing examination is 3/7 and of student B passing is 5/7 Assuming the two events “A passes”, -Maths 9th

Last Answer : p1 = P(A) = \(rac{3}{7}\), p2 = P(B) = \(rac{5}{7}\) ∴ q1 = P(\(\bar{A}\)) = 1 - P(A) = 1 - \(rac{3}{7}\) = \(rac{4}{7}\). q2 = P(\(\bar{B}\)) = 1 - P(B) = 1 - \(rac{5}{7}\) = \(rac{2}{7}\) ... passes) = p1 q2 + q1 p2 = \(rac{3}{7}\) x \(rac{2}{7}\) + \(rac{4}{7}\) x \(rac{5}{7}\) = \(rac{26}{49}.\)

PQ and RS are two equal and parallel line segments.Any points M not lying on PQ or RS is joined to Q and S and lines through P parallel to SM meet at N.Prove that line segments MN and PQ are equal and parallel to each other. -Maths 9th

Description : PQ and RS are two equal and parallel line segments.Any points M not lying on PQ or RS is joined to Q and S and lines through P parallel to SM meet at N.Prove that line segments MN and PQ are equal and parallel to each other. -Maths 9th

Last Answer : hope its clear

Let ABCD be a quadrilateral. Let X and Y be the mid-points of AC and BD respectively and the lines through X and Y respectively -Maths 9th

Description : Let ABCD be a quadrilateral. Let X and Y be the mid-points of AC and BD respectively and the lines through X and Y respectively -Maths 9th

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A line passes through the point of intersection of the lines 100x + 50y – 1 = 0 and 75x + 25y + 3 = 0 and makes equal intercepts on the axes. -Maths 9th

Description : A line passes through the point of intersection of the lines 100x + 50y – 1 = 0 and 75x + 25y + 3 = 0 and makes equal intercepts on the axes. -Maths 9th

Last Answer : (d) x + 2y = 2Let the required equation make intercept on x-axis = 2a ⇒ intercept made on y-axis = a ∴ Eqn of the given line in the intercept from:\(rac{x}{2a}+rac{y}{a}=1\) ...(i)Since the line ... 1 ⇒ a = 1.∴ Required equation of line : \(rac{x}{2 imes1}+rac{y}{1}=1\) ⇒ x + 2y = 2.

Find the equation of the line which passes through the point of intersection of the lines 2x – y + 5 = 0 -Maths 9th

Description : Find the equation of the line which passes through the point of intersection of the lines 2x – y + 5 = 0 -Maths 9th

Last Answer : (a) 45º 3x + y - 7 = 0 ⇒ y = -3x + 7 ⇒ Slope (m1) = -3 x + 2y + 9 = 0 ⇒ y = \(rac{-x}{2}\) - \(rac{9}{2}\) ⇒ Slope (m2) = \(-rac{1}{2}\)If θ is the angle between the given lines, then tan θ = \(\ ... \bigg|rac{-rac{5}{2}}{1+rac{3}{2}}\bigg|\)= \(\bigg|rac{-rac{5}{2}}{rac{5}{2}}\bigg|\) = 1∴ θ = 45°.

Whether the pair of given lines are parallel or not give reason. -Maths 9th

Description : Whether the pair of given lines are parallel or not give reason. -Maths 9th

Last Answer : Where are the lines please tell first

X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Description : X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)