The locus of a point which moves such that th | vedclass

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

Question

(D) Let the point be $P(x, y)$. The vertices of the triangle are $A(1, 2)$,$B(-2, 5)$,and $P(x, y)$.The area of the triangle is given by the formula:

Vedclass · Accepted Answer

$3x + 3y + 7 = 0 \quad \& \quad 3x + 3y - 25 = 0$

Vedclass · Answer

(D) Let the point be $P(x, y)$. The vertices of the triangle are $A(1, 2)$,$B(-2, 5)$,and $P(x, y)$.The area of the triangle is given by the formula: $	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Substituting the values: $8 = \frac{1}{2} |1(5 - y) + (-2)(y - 2) + x(2 - 5)|$.$16 = |5 - y - 2y + 4 - 3x| = |9 - 3y - 3x|$.$16 = |9 - 3(x + y)|$.This gives two cases:Case $1$: $9 - 3(x + y) = 16 \implies -3(x + y) = 7 \implies 3x + 3y + 7 = 0$.Case $2$: $9 - 3(x + y) = -16 \implies -3(x + y) = -25 \implies 3x + 3y - 25 = 0$.Thus,the locus is $3x + 3y + 7 = 0$ and $3x + 3y - 25 = 0$.

The locus of a point which moves such that the area of the triangle formed by it with the vertices $(1, 2)$ and $(-2, 5)$ is $8$ sq. units is/are

Similar Questions

The equation of the locus of the point of intersection of the straight lines $x \sin \theta + (1 - \cos \theta) y = a \sin \theta$ and $x \sin \theta - (1 + \cos \theta) y + a \sin \theta = 0$ is

$A$ variable line passes through the fixed point $(\alpha, \beta)$. The locus of the foot of the perpendicular from the origin on the line is

Find the fixed point through which the line $x(a + 2b) + y(a + 3b) = a + b$ always passes for all values of $a$ and $b$.

$A$ straight rod of length $4$ units slides such that its ends $A$ and $B$ always lie on the $X$ and $Y$-axes respectively. Then,the locus of the centroid of $\triangle OAB$ is

If a line $AB$ of length $r$ moves so that $A$ and $B$ always lie respectively on the $X$-axis and the line $y=6x$,then the locus of the mid-point of $AB$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module