The point of the form (a, – a) always lies on the line -Maths 9th

The point of the form (a, – a) always lies on the line -Maths 9th
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(d) Taking option (d), x + y = a + (-a) = a – a = 0 [since, give point is of the form (a, -a)] Hence, the point (a, – a) always lies on the line x + y = 0.
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The point of the form (a, – a) always lies on the line -Maths 9th

Description : The point of the form (a, – a) always lies on the line -Maths 9th

Last Answer : (d) Taking option (d), x + y = a + (-a) = a – a = 0 [since, give point is of the form (a, -a)] Hence, the point (a, – a) always lies on the line x + y = 0.

The point of the form (a, a) always lies on -Maths 9th

Description : The point of the form (a, a) always lies on -Maths 9th

Last Answer : (c) Since, the given point (a, a) has same value of x and y-coordinates. Therefore, the point (a, a), must be lie on the line y = x.

The point of the form (a, a) always lies on -Maths 9th

Description : The point of the form (a, a) always lies on -Maths 9th

Last Answer : (c) Since, the given point (a, a) has same value of x and y-coordinates. Therefore, the point (a, a), must be lie on the line y = x.

If y-coordinate of a point is zero, then this point always lies -Maths 9th

Description : If y-coordinate of a point is zero, then this point always lies -Maths 9th

Last Answer : (c) If y-coordinate of a point is zero, then this point always lies on X-axis. Because perpendicular distance of the point from X-axis measured along Y-axis is zero.

If y-coordinate of a point is zero, then this point always lies -Maths 9th

Description : If y-coordinate of a point is zero, then this point always lies -Maths 9th

Last Answer : (c) If y-coordinate of a point is zero, then this point always lies on X-axis. Because perpendicular distance of the point from X-axis measured along Y-axis is zero.

For what value of p the point (p, 2) lies on the line 3x + y = 11? -Maths 9th

Description : For what value of p the point (p, 2) lies on the line 3x + y = 11? -Maths 9th

Last Answer : Solution :-

Find the point which lies on the line y x = -3 having abscissa 3. -Maths 9th

Description : Find the point which lies on the line y x = -3 having abscissa 3. -Maths 9th

Last Answer : Solution :- When x=3 then y= -9,thus the point is (3,-9)

if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Description : if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Last Answer : Since complete line is AC and B is point on it. therefore, AC is divide into 2 parts AB&BC. therefore, AC=AB+BC

if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Description : if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Last Answer : AB=AC-BC BC =AC-AB AB+BC=AB HENCE PROVED

In Fig. 10.20, two circles intersects at two points A and B.AD and AC are diameters to the circles. Prove that B lies on the line A segment DC. -Maths 9th

Description : In Fig. 10.20, two circles intersects at two points A and B.AD and AC are diameters to the circles. Prove that B lies on the line A segment DC. -Maths 9th

Last Answer : Solution :- Jion AB ∠ABD = 90° (Angle in a semicircle) Similarly, ∠ABC = 90° So, ∠ABD + ∠ABC = 90° + 90° = 180° Therefore,DBC is a line i.e.,B lies on the segment DC.

A straight line is parallel to the lines 3x – y – 3 = 0 and 3x – y + 5 = 0 and lies between them. -Maths 9th

Description : A straight line is parallel to the lines 3x – y – 3 = 0 and 3x – y + 5 = 0 and lies between them. -Maths 9th

Last Answer : (c) 3x - 4y + 15 = 0 Let (m > 0) be the gradient (slope) of the required line. Then, Equation of any line through (-5, 0) having slope = m is y - 0 = m(x - (-5)) or mx - y + 5m = 0 ...(i) Its ... (rac{3}{4}\) (∵ m is +ve)∴ Required equation: y = \(rac{3}{4}\) (x + 5)⇒ 3x - 4y + 15 = 0.

The two vertices of a triangle are (2, –1), (3, 2) and the third vertex lies on the line x + y = 5. The area of the triangle is 4 units. -Maths 9th

Description : The two vertices of a triangle are (2, –1), (3, 2) and the third vertex lies on the line x + y = 5. The area of the triangle is 4 units. -Maths 9th

Last Answer : (c) (5, 0) or (1, 4) Let the third vertex of the triangle be P(a, b). Since it lies on the line x + y = 5, a + b = 5 ...(i) Also, given area of triangle formed by the points (2, -1), (3, 2) and (a, b) = 4 ... b) - (-3a + b) = 5 + 15⇒ 4a = 20 ⇒ a = 5 ⇒ b = 0. ∴ The points are (1, 4) and (5, 0).

The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Description : The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Last Answer : Given, point (2,3) lies on the line. So, the point (2, 3) is the solution of 3x - (a -1) y = 2a - 1 On putting x = 2 and y = 3 in given solution. ∴ 3 2 - (a-1) 3 = 2a - 1 ⇒ 6 - 3a + 3 = 2a - 1 ⇒ - 3a ... 2 2) 3 = 3b ⇒ 10 - 9 = 3b ⇒ 1 = 3b ⇒ 1 / 3 = b Hence, the value of b is 1 / 3.

Name the quadrant in which the point lies : -Maths 9th

Description : Name the quadrant in which the point lies : -Maths 9th

Last Answer : (a) A (1, 1) lies in the 1st quadrant, (b) B (2, 4) lies in the 1st quadrant. (c) C (-3, -10) lies in the 3rd quadrant. (d) D (-1, 2) lies in the 2nd quadrant. (e) E (1 , -1) lies in the ... lies in the 3rd quadrant. (g) G (-3, 10) lies in the 2nd quadrant. (h) H (1, -2) lies in the 4th quadrant.

Point (-3, 5) lies in the -Maths 9th

Description : Point (-3, 5) lies in the -Maths 9th

Last Answer : (b) In point (-3, 5), x-coordinate is negative and y-coordinate is positive. So, the point lies in the second quadrant.

Point (0, – 7) lies -Maths 9th

Description : Point (0, – 7) lies -Maths 9th

Last Answer : (c) In point (0, -7) x-coordinate is zero, so it lies on Y-axis and y-coordinate is negative, so the point (0, – 7) lies on the Y-axis in the negative direction.

Point (- 10,0) lies -Maths 9th

Description : Point (- 10,0) lies -Maths 9th

Last Answer : (a) In point (-10, 0) y-coordinate is zero, so it lies on X-axis and its x-coordinate is negative, so the point (-10, 0) lies on the X-axis in the negative direction.

If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Description : If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Last Answer : (d) We know that, the perpendicular distance of a point from the X-axis gives y-coordinate of that point. Here, foot of perpendicular lies on the negative direction of X-axis, so perpendicular distance can be measure in II quadrant or III quadrant. Hence, the point P has y-coordinate = 5 or -5.

The point whose ordinate is 4 and which lies on K-axis is -Maths 9th

Description : The point whose ordinate is 4 and which lies on K-axis is -Maths 9th

Last Answer : (b) Given ordinate of the point is 4 arid the point lies on Y-axis, so its abscissa is zero. Hence, the required point is (0, 4).

The point which lies on Y-axis at a distance of 5 units in the negative direction of Y-axis is -Maths 9th

Description : The point which lies on Y-axis at a distance of 5 units in the negative direction of Y-axis is -Maths 9th

Last Answer : (C) Given the point lies on X-axis this shows that its ^-coordinate is zero. Also, it is at a distance of 5 units in negative direction of X-axis, so its y-coordinate” is negative.Hence, the required point is (0, – 5).

A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Description : A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Last Answer : Given, point lies on the positive direction of X-axis, so its y-coordinate will be zero and it is at a distance of 7 units from the X-axis, so its coordinates are (7, 0). If it lies on negative ... x-coordinate will be zero and its distance from X-axis is 7 units, so its coordinates are (0, -7).

If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Description : If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Last Answer : Since, the point (x = 3, y = 4) lies on the equation 3y = ax + 7, then the equation will be , satisfied by the point. Now, put x = 3 and y = 4 in given equation, we get 3(4) = a (3)+7 ⇒ 12 = 3a+7 ⇒ 3a = 12 – 7 ⇒ 3a = 5 Hence, the value of a is 5/3.

The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Description : The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Last Answer : Given, point (2,3) lies on the line. So, the point (2, 3) is the solution of 3x - (a -1) y = 2a - 1 On putting x = 2 and y = 3 in given solution. ∴ 3 2 - (a-1) 3 = 2a - 1 ⇒ 6 - 3a + 3 = 2a - 1 ⇒ - 3a ... 2 2) 3 = 3b ⇒ 10 - 9 = 3b ⇒ 1 = 3b ⇒ 1 / 3 = b Hence, the value of b is 1 / 3.

Name the quadrant in which the point lies : -Maths 9th

Description : Name the quadrant in which the point lies : -Maths 9th

Last Answer : (a) A (1, 1) lies in the 1st quadrant, (b) B (2, 4) lies in the 1st quadrant. (c) C (-3, -10) lies in the 3rd quadrant. (d) D (-1, 2) lies in the 2nd quadrant. (e) E (1 , -1) lies in the ... lies in the 3rd quadrant. (g) G (-3, 10) lies in the 2nd quadrant. (h) H (1, -2) lies in the 4th quadrant.

Point (-3, 5) lies in the -Maths 9th

Description : Point (-3, 5) lies in the -Maths 9th

Last Answer : (b) In point (-3, 5), x-coordinate is negative and y-coordinate is positive. So, the point lies in the second quadrant.

Point (0, – 7) lies -Maths 9th

Description : Point (0, – 7) lies -Maths 9th

Last Answer : (c) In point (0, -7) x-coordinate is zero, so it lies on Y-axis and y-coordinate is negative, so the point (0, – 7) lies on the Y-axis in the negative direction.

Point (- 10,0) lies -Maths 9th

Description : Point (- 10,0) lies -Maths 9th

Last Answer : (a) In point (-10, 0) y-coordinate is zero, so it lies on X-axis and its x-coordinate is negative, so the point (-10, 0) lies on the X-axis in the negative direction.

If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Description : If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Last Answer : (d) We know that, the perpendicular distance of a point from the X-axis gives y-coordinate of that point. Here, foot of perpendicular lies on the negative direction of X-axis, so perpendicular distance can be measure in II quadrant or III quadrant. Hence, the point P has y-coordinate = 5 or -5.

The point whose ordinate is 4 and which lies on K-axis is -Maths 9th

Description : The point whose ordinate is 4 and which lies on K-axis is -Maths 9th

Last Answer : (b) Given ordinate of the point is 4 arid the point lies on Y-axis, so its abscissa is zero. Hence, the required point is (0, 4).

The point which lies on Y-axis at a distance of 5 units in the negative direction of Y-axis is -Maths 9th

Description : The point which lies on Y-axis at a distance of 5 units in the negative direction of Y-axis is -Maths 9th

Last Answer : (C) Given the point lies on X-axis this shows that its ^-coordinate is zero. Also, it is at a distance of 5 units in negative direction of X-axis, so its y-coordinate” is negative.Hence, the required point is (0, – 5).

A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Description : A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Last Answer : Given, point lies on the positive direction of X-axis, so its y-coordinate will be zero and it is at a distance of 7 units from the X-axis, so its coordinates are (7, 0). If it lies on negative ... x-coordinate will be zero and its distance from X-axis is 7 units, so its coordinates are (0, -7).

If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Description : If the point (3, 4) lies on the graph of 3y = ax + 7, then find the value of a. -Maths 9th

Last Answer : Since, the point (x = 3, y = 4) lies on the equation 3y = ax + 7, then the equation will be , satisfied by the point. Now, put x = 3 and y = 4 in given equation, we get 3(4) = a (3)+7 ⇒ 12 = 3a+7 ⇒ 3a = 12 – 7 ⇒ 3a = 5 Hence, the value of a is 5/3.

Check whether the point (a ,– a) lies on y=x–a or not. -Maths 9th

Description : Check whether the point (a ,– a) lies on y=x–a or not. -Maths 9th

Last Answer : Solution :-

If the point (2k – 3, k + 2) lies on the graph of the equation 2x + 3y +15 = 0, find the value of k. -Maths 9th

Description : If the point (2k – 3, k + 2) lies on the graph of the equation 2x + 3y +15 = 0, find the value of k. -Maths 9th

Last Answer : Solution :-

Find the point whose ordinate is –3 and which lies on y-axis. -Maths 9th

Description : Find the point whose ordinate is –3 and which lies on y-axis. -Maths 9th

Last Answer : Solution :- (0,-3)

If a point O lies between two points P and R such that PO=OR then prove that PO= 1/2PR. -Maths 9th

Description : If a point O lies between two points P and R such that PO=OR then prove that PO= 1/2PR. -Maths 9th

Last Answer : THINGS WHICH ARE COINCIDE WITH EACH OTHER ARE EQUAL TO ONE ANOTHER PO+OR=PR 2PO=PR PO=OR PO=1/2PR HENCE PROVED

Which of the following points lies on Y-axis ? -Maths 9th

Description : Which of the following points lies on Y-axis ? -Maths 9th

Last Answer : We know that, a point lies on the Y-axis, if its x-coordinate is zero. Here, x-coordinate of points C(0, 1), D(0, 0), E(0,-1) and G(0, 5) are zero. So, these points lie on Y-axis. Also ... 0) is the intersection point of both.the axes, so we can consider that it lies on Y-axis as well as on X-axis.

Which of the following points lies on Y-axis ? -Maths 9th

Description : Which of the following points lies on Y-axis ? -Maths 9th

Last Answer : We know that, a point lies on the Y-axis, if its x-coordinate is zero. Here, x-coordinate of points C(0, 1), D(0, 0), E(0,-1) and G(0, 5) are zero. So, these points lie on Y-axis. Also ... 0) is the intersection point of both.the axes, so we can consider that it lies on Y-axis as well as on X-axis.

Write the coordinates of the vertices of a rectangle whose lenght and breadth are 7 and 4 units respectively,one vertex atthe the origin,the longer side lies on the x-axis and one of the vertices lies in the third quadrant. -Maths 9th

Description : Write the coordinates of the vertices of a rectangle whose lenght and breadth are 7 and 4 units respectively,one vertex atthe the origin,the longer side lies on the x-axis and one of the vertices lies in the third quadrant. -Maths 9th

Last Answer : Solution :-

Write the coordinates of the vertices of a rectangle whose length and breadth are 6 and 3 units respectively, one vertex at the origin, the longer side lies on the y-axis and one of the vertices lies in the second quadrant. -Maths 9th

Description : Write the coordinates of the vertices of a rectangle whose length and breadth are 6 and 3 units respectively, one vertex at the origin, the longer side lies on the y-axis and one of the vertices lies in the second quadrant. -Maths 9th

Last Answer : Solution :-

For (|x-1|)/(x+2)<1, x lies in the interval -Maths 9th

Description : For (|x-1|)/(x+2)

Last Answer : answer:

If A lies in the third quadrant and 3 tan A – 4 = 0, then what is the value of 5 sin 2 A + 3 sin A + 4 cos A? -Maths 9th

Description : If A lies in the third quadrant and 3 tan A – 4 = 0, then what is the value of 5 sin 2 A + 3 sin A + 4 cos A? -Maths 9th

Last Answer : answer:

The angle A lies in the third quadrant and it satisfies the equation 4 (sin^2x + cos x) = 1. What is the measure of angle A? -Maths 9th

Description : The angle A lies in the third quadrant and it satisfies the equation 4 (sin^2x + cos x) = 1. What is the measure of angle A? -Maths 9th

Last Answer : answer:

In-centre of a triangle lies in the interior of -Maths 9th

Description : In-centre of a triangle lies in the interior of -Maths 9th

Last Answer : answer:

The graph of a linear equation in two variables is always a straight line. True/false -Maths 9th

Description : The graph of a linear equation in two variables is always a straight line. True/false -Maths 9th

Last Answer : answer:

Any point on the line y = x is of the form (a, a) -Maths 9th

Description : Any point on the line y = x is of the form (a, a) -Maths 9th

Last Answer : (a) Every point on the line y = x has same value of x-and y-coordinates i.e., x = a and y = a. Hence, (a, a) is the required form of the solution of given linear equation.

Any point on the line y = x is of the form (a, a) -Maths 9th

Description : Any point on the line y = x is of the form (a, a) -Maths 9th

Last Answer : (a) Every point on the line y = x has same value of x-and y-coordinates i.e., x = a and y = a. Hence, (a, a) is the required form of the solution of given linear equation.

ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Description : ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Last Answer : Solution: (i) In ΔACB, M is the midpoint of AB and MD || BC , D is the midpoint of AC (Converse of mid point theorem) (ii) ∠ACB = ∠ADM (Corresponding angles) also, ∠ACB = 90° , ∠ADM = 90° and MD ⊥ AC (iii ... SAS congruency] AM = CM [CPCT] also, AM = ½ AB (M is midpoint of AB) Hence, CM = MA = ½ AB

4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Description : 4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. To prove, F is the mid-point of BC. Proof, BD intersected EF at G. In ΔBAD, E is the ... point of BD and also GF || AB || DC. Thus, F is the mid point of BC (Converse of mid point theorem)

ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.