In what ratio is the line segment joining the | vedclass

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

Question

(A) Let the line $x + y = 4$ divide the line segment joining $A(-1, 1)$ and $B(5, 7)$ in the ratio $k: 1$ at point $P(x, y)$.By the section formula,th

Vedclass · Accepted Answer

$1: 2$

Vedclass · Answer

(A) Let the line $x + y = 4$ divide the line segment joining $A(-1, 1)$ and $B(5, 7)$ in the ratio $k: 1$ at point $P(x, y)$.By the section formula,the coordinates of $P$ are:$P = \left( \frac{k(5) + 1(-1)}{k + 1}, \frac{k(7) + 1(1)}{k + 1} \right) = \left( \frac{5k - 1}{k + 1}, \frac{7k + 1}{k + 1} \right)$Since $P$ lies on the line $x + y = 4$,we substitute these coordinates into the equation:$\frac{5k - 1}{k + 1} + \frac{7k + 1}{k + 1} = 4$$\frac{5k - 1 + 7k + 1}{k + 1} = 4$$\frac{12k}{k + 1} = 4$$12k = 4(k + 1)$$12k = 4k + 4$$8k = 4$$k = \frac{4}{8} = \frac{1}{2}$Thus,the ratio $k: 1$ is $1: 2$.

In what ratio is the line segment joining the points $(-1, 1)$ and $(5, 7)$ divided by the line $x + y = 4$?

Similar Questions

Find the equations of the lines parallel to the axes and passing through the point $(-2, 3)$.

If $PS$ is the median of the triangle with vertices $P(2,2)$,$Q(6,-1)$ and $R(7,3)$,then the equation of the line passing through $(1,-1)$ and parallel to $PS$ is

The equation of the line parallel to the line $2x - 3y = 1$ and passing through the midpoint of the line segment joining the points $(1, 3)$ and $(1, -7)$ is:

Find the equation of a line that makes an angle of $60^{\circ}$ with the $x$-axis and makes an intercept of $5$ units on the negative direction of the $y$-axis.

The equation of the straight line which is perpendicular to $y = x$ and passes through $(3, 2)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module