A line cuts the x-axis at A(7, 0) and the y-a | vedclass

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

Question

(C) The slope of line $AB$ is $m_{AB} = \frac{-5 - 0}{0 - 7} = \frac{5}{7}$.Since $PQ \perp AB$, the slope of line $PQ$ is $m_{PQ} = -\frac{7}{5}$.The

Vedclass · Accepted Answer

$x^{2}+y^{2}-7x+5y=0$

Vedclass · Answer

(C) The slope of line $AB$ is $m_{AB} = \frac{-5 - 0}{0 - 7} = \frac{5}{7}$.Since $PQ \perp AB$, the slope of line $PQ$ is $m_{PQ} = -\frac{7}{5}$.The equation of line $PQ$ is $y - 0 = -\frac{7}{5}(x - a)$, which simplifies to $7x + 5y = 7a$.Since $Q(0, b)$ lies on $PQ$, $5b = 7a$, so $b = \frac{7a}{5}$.$R(h, k)$ is the intersection of $AQ$ and $BP$.The equation of line $AQ$ passing through $A(7, 0)$ and $Q(0, b)$ is $\frac{x}{7} + \frac{y}{b} = 1$.The equation of line $BP$ passing through $B(0, -5)$ and $P(a, 0)$ is $\frac{x}{a} + \frac{y}{-5} = 1$.Since $R(h, k)$ lies on both lines:$(1)$ $\frac{h}{7} + \frac{k}{b} = 1$ $\Rightarrow \frac{h}{7} + \frac{5k}{7a} = 1$ $\Rightarrow ah + 5k = 7a$ $\Rightarrow a(7 - h) = 5k$ $\Rightarrow a = \frac{5k}{7 - h}$.$(2)$ $\frac{h}{a} - \frac{k}{5} = 1$ $\Rightarrow 5h - ak = 5a$ $\Rightarrow 5h = a(k + 5)$ $\Rightarrow a = \frac{5h}{k + 5}$.Equating the two expressions for $a$:$\frac{5k}{7 - h} = \frac{5h}{k + 5}$ $\Rightarrow k(k + 5) = h(7 - h)$ $\Rightarrow k^2 + 5k = 7h - h^2$.Rearranging gives $h^2 + k^2 - 7h + 5k = 0$.Replacing $(h, k)$ with $(x, y)$, the locus is $x^2 + y^2 - 7x + 5y = 0$.

$A$ line cuts the $x$-axis at $A(7, 0)$ and the $y$-axis at $B(0, -5)$. $A$ variable line $PQ$ is drawn perpendicular to $AB$ cutting the $x$-axis at $P(a, 0)$ and the $y$-axis at $Q(0, b)$. If $AQ$ and $BP$ intersect at $R(h, k)$, the locus of $R$ is

Similar Questions

If the point $(x, y)$ is equidistant from the points $(a + b, b - a)$ and $(a - b, a + b)$,then:

Suppose that the three points $A, B$ and $C$ in the plane are such that their $x$-coordinates as well as $y$-coordinates are in $GP$ with the same common ratio. Then,the points $A, B$ and $C$

Let $A$ be the set of all points $(\alpha, \beta)$ such that the area of the triangle formed by the points $(5, 6), (3, 2),$ and $(\alpha, \beta)$ is $12 \text{ square units}.$ Then the least possible length of a line segment joining the origin to a point in $A$ is:

Given $A(1, 1)$ and $AB$ is any line through it cutting the $x-$ axis in $B$. If $AC$ is perpendicular to $AB$ and meets the $y-$ axis in $C$,then the equation of the locus of the midpoint $P$ of $BC$ is

If the line $\frac{x}{a} + \frac{y}{b} = 1$ is a variable line such that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$,then the locus of the foot of the perpendicular from the origin to the line is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module