the points A,Band C are(4,0),(2,2),and(0,6) respectively AB produced cuts the y-axis at Pand CB produced cuts the x-axis at Q - Maths - Straight Lines

Question

the points A,Band C are(4,0),(2,2),and(0,6) respectively AB
produced cuts the y-axis at Pand CB produced cuts the x-axis at Q
find the co-ordinates of the pointsP and q Find the eq of the
straight line joining the mid points of AC nad OB( Where O is the
origin )and verify that this line passes through the mid point of
PQ

Manmeet Singh · Accepted Answer

We have,

Slope of BC = slope of QB $\frac{2 - 6}{2 - 0} = \frac{2 - 0}{2 - x} \Rightarrow \frac{- 4}{2} = \frac{2}{2 - x} \Rightarrow - 8 + 4 x = 4 \Rightarrow 4 x = 12 \Rightarrow x = 3 T h e c o o r d i n a t e s o f p o int Q = (3, 0) S i m i l a r l y, s l o p e o f A B = s l o p e o f B P \Rightarrow \frac{2 - 0}{2 - 4} = \frac{2 - y}{2 - 0} \Rightarrow \frac{2}{- 2} = \frac{2 - y}{2} \Rightarrow - 1 = \frac{2 - y}{2} \Rightarrow - 2 = 2 - y \Rightarrow y = 4 T h e c o o r d i n a t e s o f P a r e (0, 4)$ Mid point of AC can be calculated by using midpoint formula, Let the midpoint be R $R = (\frac{4 + 0}{2}, \frac{0 + 6}{2}) R = (2, 3)$ Similarly, let the midpoint of OB be S $S = (\frac{0 + 2}{2}, \frac{0 + 2}{2}) S = (1, 1)$ The slope of line joining points R and S = $\frac{3 - 1}{2 - 1} = 2$ The equation of line joining points R and S is, $y - y_{1} = m (x - x_{1}) \Rightarrow y - 1 = 2 (x - 1) \Rightarrow y - 1 = 2 x - 2 \Rightarrow 2 x - y = 1 N o w t h e m i d p o int o f P Q = (\frac{0 + 3}{2}, \frac{4 + 0}{2}) = (\frac{3}{2}, 2) T h i s p o int s a t i s f i e s t h e e q u a t i o n o f l i n e j o i n i n g t h e m i d p o int s o f A C a n d O B 2 (\frac{3}{2}) - 2 = 3 - 2 = 1 = R H S$