Question

3. Find the equation of a straight line whose
(i) slope is -3 and y-intercept is 4.
(ii) angle of inclination is 600
and y-intercept is 3.

Answer 1

Hey there!

Straight line : It is a curve, whose any two points are joined to form line segments, and those lie wholly on it.

The general form of a straight line : ax + by + c = 0 where, | a | + | b | ≠ 0 that is a & b shouldn't be 0 at same time.

The equation of a straight line with given slope m and y - intercept " c " is y = mx + c

The equation of a straight line with given slope m and x - intercept " c" is y = m ( x - c) .

Also, The slope of a line is the inclination of the line with respect to X - axis. m = tanθ
Now,
We have to find equation of a straight line whose
( i) Slope is -3 , y - intercept is 4 .

So,
m = -3
c = 4

We know that,
With given y - intercept and slope, Equation of straight line
$\boxed{ y = mx + c }$

So, y = ( -3 ) x + 4
=> y = -3x + 4
=> 3x + y - 4 = 0

The equation of line slope is -3 and y-intercept is 4 is " 3x + y - 4 = 0 "

Again,
We have to find the equation of straight whose inclination is 600° , y intercept is 3 .

We know that,
$\boxed { m = tan \theta }$

So, m = tan600 = tan ( 90* 6 + 60 ) = tan60 = √3 .

Hence, We have m = √3 , c = 3 .

The equation of straight line ,
y = mx + c
y = √3 x + 3
√3x - y + 3 = 0
3x - √3y + 3√3 = 0 .

Therefore, the required straight lines are 3x + y - 4 = 0 , 3x - √3y + 3√3 = 0

Hope helped!

Answer 2

Solution :

*******************************************

Equation of a line whose ,

slope = m , y-intercept = c is

y = mx + c

*****************************************
Here ,

i ) slope ( m ) = -3 ,

y-intercept = c = 4

Required equation ,

y = -3x + 4

ii ) inclination ( x ) = 60°

slope ( m ) = tan x

m = tan 60°

=> m = √3

slope = m = √3 ,

y-intercept ( c ) = 3

Required equation ,

y = √3x + 3

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