The coordinates of the vertices A and B of an | vedclass

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

Question

(B) Let $C$ be $(\alpha, \beta)$.Since $AC = BC$,we have $AC^2 = BC^2$.$(\alpha + 2)^2 + (\beta - 3)^2 = (\alpha - 2)^2 + (\beta - 0)^2$$\alpha^2 + 4\

Vedclass · Accepted Answer

$\left(1, \frac{17}{6}\right)$

Vedclass · Answer

(B) Let $C$ be $(\alpha, \beta)$.Since $AC = BC$,we have $AC^2 = BC^2$.$(\alpha + 2)^2 + (\beta - 3)^2 = (\alpha - 2)^2 + (\beta - 0)^2$$\alpha^2 + 4\alpha + 4 + \beta^2 - 6\beta + 9 = \alpha^2 - 4\alpha + 4 + \beta^2$$8\alpha - 6\beta + 9 = 0$  ......$(1)$Slope of $AB = \frac{0 - 3}{2 - (-2)} = -\frac{3}{4}$.The equation of a line parallel to $AB$ is $y = -\frac{3}{4}x + c$.Given $y$-intercept $c = \frac{43}{12}$,the equation is $y = -\frac{3}{4}x + \frac{43}{12}$,which simplifies to $9x + 12y = 43$.Since this line passes through $C(\alpha, \beta)$,we have $9\alpha + 12\beta = 43$ ......$(2)$.Solving equations $(1)$ and $(2)$:From $(1)$,$6\beta = 8\alpha + 9 \Rightarrow 12\beta = 16\alpha + 18$.Substituting into $(2)$: $9\alpha + 16\alpha + 18 = 43$ $\Rightarrow 25\alpha = 25$ $\Rightarrow \alpha = 1$.Then $12\beta = 43 - 9(1) = 34 \Rightarrow \beta = \frac{34}{12} = \frac{17}{6}$.Thus,$C = \left(1, \frac{17}{6}\right)$.

The coordinates of the vertices $A$ and $B$ of an isosceles triangle $ABC$ $(AC = BC)$ are $(-2, 3)$ and $(2, 0)$ respectively. $A$ line parallel to $AB$ and having a $y$-intercept equal to $\frac{43}{12}$ passes through $C$. Then the coordinates of $C$ are:

Similar Questions

Let $A(-1, 1)$,$B(3, 4)$,and $C(2, 0)$ be three given points. $A$ line $y = mx$,$m > 0$,intersects lines $AC$ and $BC$ at points $P$ and $Q$ respectively. Let $A_1$ and $A_2$ be the areas of $\Delta ABC$ and $\Delta PQC$ respectively,such that $A_1 = 3A_2$. Then the value of $m$ is equal to:

The equation of the line passing through the point of intersection of the lines $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$ and making equal non-zero intercepts on the coordinate axes 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 the equation of the straight line passing through the point of intersection of $x+2y-19=0$ and $x-2y-3=0$ and which is at a perpendicular distance of $5$ units from the point $(-2,4)$ is $5x+by+c=0$,then a possible value of $5+b+c$ is

Three lines $x + 2y + 3 = 0$,$x + 2y - 7 = 0$,and $2x - y - 4 = 0$ form three sides of two squares. Find the equation of the fourth side of each square.

Vedclass Test Series

Exam Paper Generator

Online Exam Module