Question

7. If the straight line passing through the points (h, 3) and (4, 1) intersects the line 7x-9y-19=0 at right angle, then find the value of h

Answer 1

HELLO DEAR,

WE KNOW THAT:-

→Slope of a line passing through the points$(x_1,\:\:y_1)$and$(x_2,\:\:y_2)$ is
$\mathbf{m=\frac{y_2 - y_1}{x_2-x_1}}$

→If two lines are perpendicular, then the product of their slopes is -1.

NOW,

Slope of the line passing through (h,3) and (4,1) is
$\mathbf{m_1=\frac{1-3}{4-h} =\frac{-2}{4-h}}$

Equation of the given line is 7x-9y-19=0

9y = 7x - 19

$\mathbf{x = \frac{7}{9}x - \frac{19}{9}}$

Hence,

the slope is 7/9

$\mathbf{m_2=\frac{7}{9}}$

NOW, the lines are also perpendicular,

the product of the slopes is -1.

$\mathbf{m_1*m_2 = -1}$

$\mathbf{\frac{-2}{4-h}* \frac{7}{9}=-1}$

$\mathbf{\frac{-14}{36 - 9h} = -1}$

$\mathbf{-14 = -36 + 9h}\\ \\ \mathbf{22 = 9h}\\ \\ \mathbf{h = \frac{22}{9}}$

$\large{\mathbf{\underline{I\:\: HOPE\:\: IT'S\:\: HELP\:\:YOU\:\: DEAR,\:\: THANKS}}}$

Answer 2

Solution:

i ) Slope of line passing through

the points A( h , 3 ) = ( x1 , y1 )

and B( 4 , 1 ) = ( x2 , y2 ) is

m1 = ( y2 - y1 )/( x2 - x1 )

=> m1 = ( 1 - 3 )/( 4 - h )

=> m1 = -2/( 4 - h ) ----( 1 )

ii ) Slope of a line 7x - 9y - 19 = 0 is

m2 = - ( a/b )

=> m2 = -[7/(-9 )]

=> m2 = 7/9 ---( 2 )

According to the problem given ,

m1 × m2 = -1

( -2 )/( 4 - h ) × 7/9 = -1

=> -14= -9 ( 4 - h )

=> -14 + 9( 4 - h ) = 0

=> -14 + 36 - 9h = 0

=> 22 - 9h = 0

=> 9h = 22

=> h = 22/9

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