The line joining the points $A (0, 5)$ and $B (4, 2)$ is perpendicular to the line joining the points $C (- 1, - 2)$ and $D (5, b)$ . Find the value of $b$ .

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Solution

Consider the given points.
$A (0, 5)$ and $B (4, 2)$

Slope of $A B = m_{1} = \frac{2 - 5}{4 - 0} = \frac{- 3}{4}$

Consider the given other points.
$C (- 1, - 2)$ and $D (5, b)$

Slope of $C D = m_{2} = \frac{b + 2}{5 + 1} = \frac{b + 2}{6}$

Since, both lines are perpendicular to each other so,
$m_{1} . m_{2} = - 1$

Therefore,
$\frac{- 3}{4} \times \frac{b + 2}{6} = - 1$

$\frac{1}{4} \times \frac{b + 2}{2} = 1$

$b + 2 = 8$

$b = 6$

Hence, this is the answer.

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The line joining the points A(0,5) and B(4,2) is perpendicular to the line joining the points C(−1,−2) and D(5,b). Find the value of b.

Consider the given points.A(0,5) and B(4,2)Slope of AB=m1=2−54−0=−34Consider the given other points.C(−1,−2) and D(5,b)Slope of CD=m2=b+25+1=b+26Since, both lines are perpendicular to each other so,m1.m2=−1Therefore,−34×b+26=−114×b+22=1b+2=8b=6Hence, this is the answer.

The line joining the points $A (0, 5)$ and $B (4, 2)$ is perpendicular to the line joining the points $C (- 1, - 2)$ and $D (5, b)$ . Find the value of $b$ .