A line,with a slope greater than 1,passes thr | vedclass

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

Question

(C) Let the point $B$ be $(x_1, y_1)$. Since $B$ lies on $x - y - 2 = 0$,we have $y_1 = x_1 - 2$,so $B = (x_1, x_1 - 2)$.The distance $AB = \sqrt{(x_1

Vedclass · Accepted Answer

$x + 2y = 6$

Vedclass · Answer

(C) Let the point $B$ be $(x_1, y_1)$. Since $B$ lies on $x - y - 2 = 0$,we have $y_1 = x_1 - 2$,so $B = (x_1, x_1 - 2)$.The distance $AB = \sqrt{(x_1 - 4)^2 + (x_1 - 2 - 3)^2} = \frac{\sqrt{29}}{3}$.Squaring both sides: $(x_1 - 4)^2 + (x_1 - 5)^2 = \frac{29}{9}$.$x_1^2 - 8x_1 + 16 + x_1^2 - 10x_1 + 25 = \frac{29}{9} \implies 2x_1^2 - 18x_1 + 41 = \frac{29}{9}$.$18x_1^2 - 162x_1 + 369 = 29 \implies 18x_1^2 - 162x_1 + 340 = 0 \implies 9x_1^2 - 81x_1 + 170 = 0$.Solving the quadratic equation: $(3x_1 - 10)(3x_1 - 17) = 0$,so $x_1 = \frac{10}{3}$ or $x_1 = \frac{17}{3}$.If $x_1 = \frac{10}{3}$,then $y_1 = \frac{10}{3} - 2 = \frac{4}{3}$. The slope $m = \frac{4/3 - 3}{10/3 - 4} = \frac{-5/3}{-2/3} = 2.5 > 1$.If $x_1 = \frac{17}{3}$,then $y_1 = \frac{17}{3} - 2 = \frac{11}{3}$. The slope $m = \frac{11/3 - 3}{17/3 - 4} = \frac{2/3}{5/3} = 0.4 Since the slope must be greater than $1$,we take $B = (\frac{10}{3}, \frac{4}{3})$.Checking the options for $(\frac{10}{3}, \frac{4}{3})$: $x + 2y = \frac{10}{3} + 2(\frac{4}{3}) = \frac{10+8}{3} = 6$. Thus,$B$ lies on $x + 2y = 6$.

$A$ line,with a slope greater than $1$,passes through the point $A(4, 3)$ and intersects the line $x - y - 2 = 0$ at the point $B$. If the length of the line segment $AB$ is $\frac{\sqrt{29}}{3}$,then $B$ also lies on the line:

Similar Questions

If the equation of the base of an equilateral triangle is $2x - y = 1$ and the vertex is $(-1, 2)$,then the length of the side of the triangle is

The point on the line $4x - y - 2 = 0$ which is equidistant from the points $(-5, 6)$ and $(3, 2)$ is:

Let the lines $3x - 4y - \alpha = 0$,$8x - 11y - 33 = 0$,and $2x - 3y + \lambda = 0$ be concurrent. If the image of the point $(1, 2)$ in the line $2x - 3y + \lambda = 0$ is $\left(\frac{57}{13}, \frac{-40}{13}\right)$,then $|\alpha \lambda|$ is equal to:

If a straight line $L$ perpendicular to the line $3x - 4y = 6$ forms a triangle of area $6$ square units with the coordinate axes,then the minimum perpendicular distance from the point $(1, 1)$ to the line $L$ is

Two non-parallel sides of a rhombus are parallel to the lines $x+y-1=0$ and $7x-y-5=0$. If $(1,3)$ is the centre of the rhombus and one of its vertices $A(\alpha, \beta)$ lies on $15x-5y=6$,then one of the possible values of $(\alpha+\beta)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module