A point on the line y-x whose perpendicular distance from the line x-4-y-3-1 is 4- has the coordinates

The correct options are A ( 8/7, 8/7 ) B ( 32/7,32/7 ) Given equation is 3x+4y 12=0 let's take point is (a,a) on y=x from above form ...

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Standard XII

Mathematics

Secant of a Curve y =f(x)

A point on th...

Question

A point on the line $y = x$ whose perpendicular distance from the line $\frac{x}{4} + \frac{y}{3} = 1$ is 4, has the coordinates

(- \frac{8}{7}, - \frac{8}{7})

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(\frac{32}{7}, \frac{32}{7})

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(\frac{3}{2}, \frac{3}{2})

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none of these

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Solution

The correct options are
A $(- \frac{8}{7}, - \frac{8}{7})$
B $(\frac{32}{7}, \frac{32}{7})$
Given equation is $3 x + 4 y - 12 = 0$
let's take point is $(a, a)$ on $y = x$
from above formula

$∣ ∣ ∣ ∣ ∣ \frac{3 a + 4 a - 12}{\sqrt{(3^{2} + (4^{2}))}} ∣ ∣ ∣ ∣ ∣ = 4$
$\frac{3 a + 4 a - 12}{\sqrt{(3^{2} + (4^{2}))}} = \pm 4$
$a = \frac{- 8}{7}, \frac{32}{7}$
So points the are
$(\frac{- 8}{7}, \frac{- 8}{7}), (\frac{32}{7}, \frac{32}{7})$

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A point on the line y=x whose perpendicular distance from the line x4+y3=1 is 4, has the coordinates

The correct options are A (−87,−87) B (327,327)Given equation is 3x+4y−12=0let's take point is (a,a) on y=xfrom above formula ∣∣ ∣ ∣∣3a+4a−12√(32+(42))∣∣ ∣ ∣∣=43a+4a−12√(32+(42))=±4a=−87,327So points the are(−87,−87),(327,327)

A point on the line $y = x$ whose perpendicular distance from the line $\frac{x}{4} + \frac{y}{3} = 1$ is 4, has the coordinates