A straight line L cuts both the lines 5x - y | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the line $L$ pass through the point $P(1, 5)$. Let the line $L$ intersect $L_1: 5x - y - 4 = 0$ at $A(x_1, y_1)$ and $L_2: 3x + 4y - 4 = 0$ at

Vedclass · Accepted Answer

$83x - 35y + 92 = 0$

Vedclass · Answer

(D) Let the line $L$ pass through the point $P(1, 5)$. Let the line $L$ intersect $L_1: 5x - y - 4 = 0$ at $A(x_1, y_1)$ and $L_2: 3x + 4y - 4 = 0$ at $B(x_2, y_2)$.Since $P(1, 5)$ is the midpoint of $AB$,we have $x_1 + x_2 = 2$ and $y_1 + y_2 = 10$.Since $A$ lies on $L_1$,$5x_1 - y_1 - 4 = 0$,so $y_1 = 5x_1 - 4$.Since $B$ lies on $L_2$,$3x_2 + 4y_2 - 4 = 0$,so $y_2 = \frac{4 - 3x_2}{4} = 1 - \frac{3}{4}x_2$.Substituting $x_2 = 2 - x_1$ and $y_2 = 10 - y_1$ into the equation for $B$:$3(2 - x_1) + 4(10 - y_1) - 4 = 0 \implies 6 - 3x_1 + 40 - 4y_1 - 4 = 0 \implies 3x_1 + 4y_1 = 42$.Now solve the system: $y_1 = 5x_1 - 4$ and $3x_1 + 4y_1 = 42$.$3x_1 + 4(5x_1 - 4) = 42 \implies 3x_1 + 20x_1 - 16 = 42 \implies 23x_1 = 58 \implies x_1 = \frac{58}{23}$.Then $y_1 = 5(\frac{58}{23}) - 4 = \frac{290 - 92}{23} = \frac{198}{23}$.The slope $m$ of line $L$ passing through $A(\frac{58}{23}, \frac{198}{23})$ and $P(1, 5)$ is:$m = \frac{5 - \frac{198}{23}}{1 - \frac{58}{23}} = \frac{\frac{115 - 198}{23}}{\frac{23 - 58}{23}} = \frac{-83}{-35} = \frac{83}{35}$.The equation of $L$ is $y - 5 = \frac{83}{35}(x - 1) \implies 35y - 175 = 83x - 83 \implies 83x - 35y + 92 = 0$.

$A$ straight line $L$ cuts both the lines $5x - y - 4 = 0$ and $3x + 4y - 4 = 0$. The segment of $L$ between the two lines is bisected at the point $(1, 5)$. The equation of $L$ is

Similar Questions

Consider the triangles with vertices $A(2,1)$,$B(0,0)$,and $C(t,4)$,where $t \in [0,4]$. If the maximum and the minimum perimeters of such triangles are obtained at $t=\alpha$ and $t=\beta$ respectively,then $6\alpha + 21\beta$ is equal to $.........$.

If the point $\left(\alpha, \frac{7 \sqrt{3}}{3}\right)$ lies on the curve traced by the mid-points of the line segments of the lines $x \cos \theta + y \sin \theta = 7, \theta \in \left(0, \frac{\pi}{2}\right)$ between the coordinate axes,then $\alpha$ is equal to

If the equation of a curve remains unchanged by replacing $x$ with $y$ and $y$ with $x$,then the curve is

The equation of the locus of a point which is equidistant from the points $(2, 3)$ and $(4, 5)$ is

If $A = (a, 0)$ and $B = (-a, 0)$,then the locus of a point $P = (x, y)$ such that $PA^2 - PB^2 = a^2$ is.

Vedclass Test Series

Exam Paper Generator

Online Exam Module