Question

Show that lines x - 2y - 7 = 0 and
2x + y + 1 = 0 are perpendicular to each
other. Find their point of intersection.​

Answer 1

Answer:

point of intersection (1,-3)

Step-by-step explanation:

To find whether lines x - 2y - 7 = 0 and

2x + y + 1 = 0 are perpendicular to each

other.

we know that two lines are perpendicular if their slope

$\boxed{m_1m_2 = - 1}$

we know that standard equation of line

$y = mx + c$

For line 1

$x - 2y - 7 = 0 \\ \\ - 2y = - x + 7 \\ \\ y = \frac{1}{2} x - \frac{7}{2} \\ \\ m_1 = \frac{1}{2}$

For line 2

$2x + y + 1 = 0 \\ \\ y = - 2x - 1 \\ \\ m_2 = - 2$

from both slopes

$= m_1m_2 \\ \\ = (\frac{1}{2} )( - 2) \\ \\ m_1m_2 = - 1$

Hence both lines are perpendicular.

To find the intersecting points of the two lines

$2x - 4y = 14 \\ </p><p>2x + y = - 1 \\ ( - ) \: \: ( - ) \: \: ( + ) \\ - - - - - - - \\ - 5y = 15 \\ \\ y = - 3 \\ \\ put \: the \: value \: of \: y \: in \: eq \\ 2x + ( - 3) = - 1 \\ \\ 2x = - 1 + 3 \\ \\ 2x = 2 \\ \\ x = 1$

Hence (1,-3) is the intersecting point.

Hope it helps you.

Answer 2

Answer:

point of intersection is (x=1, y=-2)

Step-by-step explanation:

The slopes of perpendicular lines are opposite reciprocals of each other.

To determine the slopes of the lines, we express them in standard form

ie y=mx+c

so x-2y-7=0 becomes  

$x-2y-7=0\\2y=x-7\\y=\frac{1}{2}x-\frac{7}{2}$

its slope is $\frac{1}{2}$

line 2x+y+1=0 becomes

$2x+y+1=0\\y=-x-1$

its slope is -2

To get the opposite reciprocal of a number we first get the reciprocal by dividing 1 by the number, then its opposite by multiplying this by -1.

so for -2, its reciprocal is $-\frac{1}{2}$ and the opposite of this is $-\frac{1}{2}*-1=\frac{1}{2}$

Hence proven.

At the point of intersection, values of y are equal

i.e

$\frac{1}{2}x-\frac{7}{2}=y=-2x-1\\\frac{1}{2}x+2x=-1+\frac{7}{2}\\ \frac{5}{2}x=\frac{5}{2}\\  x=1$

The value of y is

$y=-x-1\\y=-1-1\\y=-2$

The point of intersection is (x=1, y=-2)

i.e.

(1,-2)