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Last Answer : (d) \(\lambda\) = \(\mu\) Let the equation of the line in the intercept from be\(rac{x}{\lambda}\)+ \(rac{y}{\mu}\) = 1Since it passes through (4, 3) and (2, 5)\(rac{4}{\lambda}\) + \(rac{3}{\mu}\) = 1 ... ) = 1 - \(rac{3}{7}\) = \(rac{4}{7}\) = \(\lambda\) = 7∴ \(\lambda\) = \(\mu\) = 7.

Description : Why can a function have multiple x-intercepts but only one y-intercept?

Last Answer : It can, but only if the line drawn is a curve. ---------------------------------------------------------------------------------- The function is a polynomial of the form: y = f(x) = ... value, eg y = x² has y = 4 for x =  2). Thus there will only be one y value for x = 0, ie only one y-intecept.

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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

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

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.

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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.

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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.

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Description : The graph of the linear equation y = x passes through the point. -Maths 9th

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Description : The graph of the linear equation y = x passes through the point. -Maths 9th

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Last Answer : Solution :-

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Last Answer : hope it helps

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Description : The graph of y = 6 is a Line -Maths 9th

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