Question

Find the equation of a line perpendicular to the line x - 2y + 3 = 0 and passing through the point (1,-2).

Answer 1

Equation of line is $\bf \red{2x+y=0}$.

Given:

• Equation of a line x-2y+3=0
• A point (1,-2).

To find:

• Find the equation of a line perpendicular to the line x - 2y + 3 = 0 and passing through the point (1,-2).

Solution:

Formula/Concept to be used:

• Equation of a line passing through a point (x1,y1):$\bf (x - x_1) = m(y - y_1)$
• Slope of two perpendicular lines:If slope of two perpendicular lines are m1 and m2 then $\bf m_1m_2 = - 1$

Step 1:

Find slope of line.

To find the slope of line which is perpendicular to the given line.

Write the given equation in slope intercept form.

$x - 2y + 3 = 0$

or

$- 2y = - x - 3$

or

$\bf y = \frac{1}{2} x + \frac{3}{2}$

As slope intercept form is

$\boxed{\bf y = mx + c }$

So, on comparing ,it is clear that slope is 1/2.

Let , it is $m_1$,

$m_1=\frac{1}{2}$

Now, let slope of other line is $m_2$

So,

$m_2 = \frac{ - 1}{m_1}$

or

$\bf m_2 = - 2$

Step 2:

Find the equation of line.

Line is passing through (1,-2), it's slope is -2.

So,

Equation of line

$y - ( - 2) = - 2(x - 1)$

or

$y + 2 = - 2x + 2$

or

$\bf 2x + y = 0$

Thus,

Equation of line is 2x+y=0.

#SPJ3

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Answer 2

We have to find the line perpendicular to the line x-2y+3=0 and passing through the point (1,-2).

Answer:

Given, line is x-2y+3=0.

Let's change the line into the form y= mx+c

Changing the given line into slope form, will helps us to find the value of slope.

Therefore,

$x - 2y + 3 = 0 \\ 2y = x + 3 \\ y = (1 \div 2)x + 3 \div 2$

It is of the form y= mx+c, where m= 1/2

Now, to find the line perpendicular to the given line, we need to find the slope of perpendicular line.

Slope of perpendicular line is

$- (1 \div m) \\ = - (1 \div (1 \div 2)) \\ = - 2$

Therefore, the slope of the perpendicular line is -2.

To find the equation of the line with slope m= -2 we need to substitute the value of m in y=mx+c

Therefore, equation of line having slope m= -2 is,

$y = - 2 \times x + c$

Let us suppose that it is equation 1.

In the question, given that the perpendicular line is passing through the point (1,-2)

By substituting the point in the equation we get the value of c.

$- 2 = - 2 \times 1 + c \\ c = - 2 + 2 \\ c = 0$

Therefore the value of c is 0.

Substitute the value of c in equation 1.

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

Therefore the equation of perpendicular line to x-2y+3=0 which is passing through the point (1,-2) is

2x+y=0.

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