Question 25

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Jambmaths question: 

Find the value of p if the line joining (p, 4) and (6, –2) is perpendicular to the line joining (2, p) and (–1, – 3)

Option A: 

4

Option B: 

6

Option C: 

3

Option D: 

0

Jamb Maths Solution: 

For two lines to be perpendicular, the product of the two slopes is equal to –1 (i.e. m1 m2   = –1)$\begin{align} & {{l}_{1}}: \\ & ({{x}_{1}},{{y}_{1}})=(1,4) \\ & ({{x}_{2}},{{y}_{2}})=(6,-2) \\ & {{m}_{1}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\frac{-2-4}{6-p} \\ & {{m}_{1}}=\frac{-6}{6-p}----(i) \\ & {{l}_{2}} \\ & ({{x}_{1}},{{y}_{1}})=(2,p) \\ & ({{x}_{2}},{{y}_{2}})=(-1,-3) \\ & {{m}_{2}}=\frac{-3-p}{-1-2} \\ & {{m}_{2}}=\frac{-3-p}{-3}---(ii) \\ & \text{Since }{{l}_{1}}\bot {{l}_{2}} \\ & {{m}_{1}}\times {{m}_{2}}=-1 \\ & \frac{(-3-p)}{-3}\times \frac{(-6)}{6-p}=-1 \\ & \frac{(-3-p)(2)}{6-p}=-1 \\ & -6-2p=p-6 \\ & 3p=-6+6 \\ & p=0 \\\end{align}$

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