Given two points Q(3,4) and R(1,2). What is t | vedclass

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

Question

(C) The condition $PQ+PR=QR$ implies that the point $P$ must lie on the line segment $QR$. First,we find the equation of the line passing through $Q(3

Vedclass · Accepted Answer

$(2,3)$

Vedclass · Answer

(C) The condition $PQ+PR=QR$ implies that the point $P$ must lie on the line segment $QR$. First,we find the equation of the line passing through $Q(3,4)$ and $R(1,2)$. The slope $m = \frac{2-4}{1-3} = \frac{-2}{-2} = 1$. The equation of the line $QR$ is $y - 2 = 1(x - 1)$,which simplifies to $x - y + 1 = 0$. Since $P$ lies on both the line $2x - y - 1 = 0$ and the line $x - y + 1 = 0$,we solve the system of equations: $2x - y = 1$ $x - y = -1$ Subtracting the second equation from the first gives $(2x - x) - (y - y) = 1 - (-1)$,so $x = 2$. Substituting $x = 2$ into $x - y = -1$,we get $2 - y = -1$,which implies $y = 3$. Thus,the point $P$ is $(2,3)$.

Given two points $Q(3,4)$ and $R(1,2)$. What is the point $P(x, y)$ on the line $2x-y-1=0$ for which $PQ+PR=QR$ holds?

Similar Questions

For the straight lines given by the equation $(2 + k)x + (1 + k)y = 5 + 7k$,for different values of $k$,which of the following statements is true?

If the lines $x + 2ay + a = 0$,$x + 3by + b = 0$ and $x + 4cy + c = 0$ are concurrent,then $a$,$b$ and $c$ are in

If $3a + 5b + 6c = 0$,then the family of lines $ax + by + c = 0$ passes through the fixed point:

What is the nature of the points $(-a, -b)$,$(0, 0)$,$(a, b)$,and $(a^2, ab)$?

The equation of a line passing through the point of intersection of the lines $x - 2y + 8 = 0$ and $3x - y + 4 = 0$ and passing through the origin is

Vedclass Test Series

Exam Paper Generator

Online Exam Module