Let area of a triangle PQR with vertices P (5,4), Q (-2,4) and R(a, b) be 35 square units. If its or

English

Gujarati

Hindi

9.Straight Line

medium

Let the area of a $\triangle PQR$ with vertices $P (5,4), Q (-2,4)$ and $R(a, b)$ be $35$ square units. If its orthocenter and centroid are $O\left(2, \frac{14}{5}\right)$ and $C(c, d)$ respectively, then $c+2 d$ is equal to :

A$\frac{7}{3}$

B$3$

C$2$

D$\frac{8}{3}$

(JEE MAIN-2025)

Solution

image
$\text { Equation of lines } QR =5 x +2 y +2=0$
$\text { Equation of lines } PR =10 x -3 y -38=0$
$\therefore \text { Point } R (2,-6)$
$\text { Centroid }=\left(\frac{5-2+2}{3}, \frac{4+4-6}{3}\right)$
$=\left(\frac{5}{3}, \frac{2}{3}\right)$
$c +2 d=\frac{5}{3}+\frac{4}{3}=3$

Standard 11

Mathematics

The number of possible straight lines , passing through $(2, 3)$ and forming a triangle with coordinate axes, whose area is $12 \,sq$. units , is

normal

The equation to the sides of a triangle are $x – 3y = 0$, $4x + 3y = 5$ and $3x + y = 0$. The line $3x – 4y = 0$ passes through

medium

Without using the Pythagoras theorem, show that the points $(4,4),(3,5)$ and $(-1,-1)$ are vertices of a right angled triangle.

medium

Area of the parallelogram whose sides are $x\cos \alpha + y\sin \alpha = p$ $x\cos \alpha + y\sin \alpha = q,\,\,$ $x\cos \beta + y\sin \beta = r$ and $x\cos \beta + y\sin \beta = s$ is

hard

The equations of two equal sides of an isosceles triangle are $7x – y + 3 = 0$ and $x + y – 3 = 0$ and the third side passes through the point $(1, -10)$. The equation of the third side is

hard

(IIT-1984)

Let the area of a $\triangle PQR$ with vertices $P (5,4), Q (-2,4)$ and $R(a, b)$ be $35$ square units. If its orthocenter and centroid are $O\left(2, \frac{14}{5}\right)$ and $C(c, d)$ respectively, then $c+2 d$ is equal to :

Solution

Similar Questions

The number of possible straight lines , passing through $(2, 3)$ and forming a triangle with coordinate axes, whose area is $12 \,sq$. units , is

The equation to the sides of a triangle are $x – 3y = 0$, $4x + 3y = 5$ and $3x + y = 0$. The line $3x – 4y = 0$ passes through

Without using the Pythagoras theorem, show that the points $(4,4),(3,5)$ and $(-1,-1)$ are vertices of a right angled triangle.

Area of the parallelogram whose sides are $x\cos \alpha + y\sin \alpha = p$ $x\cos \alpha + y\sin \alpha = q,\,\,$ $x\cos \beta + y\sin \beta = r$ and $x\cos \beta + y\sin \beta = s$ is

The equations of two equal sides of an isosceles triangle are $7x – y + 3 = 0$ and $x + y – 3 = 0$ and the third side passes through the point $(1, -10)$. The equation of the third side is

Start a Free Trial Now