Straight Line Part 4 Quiz

Slope of the line with inclination 60º is m = tan 60º
or, m = √3

Slope of the line with inclination 45º is m = tan 45º
or, m = 1
Equation of the line: y = (1)(x) + (-3)
or, y = x - 3

Representing the line in the y = mx + c format we get y = x - 7/3
So slope m = 1, inclination tan θ = 1,
or, inclination θ = 45º
Intercept c = -7/3

Given m = 1/2
(3 - k)/(1 - 3k - 6) = 1/2
or, 6 - 2k = 1 - 3k - 6
or, k = -11

Slope of the given line is m = 2
Equation of the line passing throught (4, -4) is:
(y + 4) = 2(x - 4)
or, y = 2x - 12
Comparing with the give equation we get
2p = -12
or, p = -6

Let the intercepts on x and y axis be 'a'
So equation is (x/a) + (y/a) = 1
Since it passes through (3, 2) it should satisfy the equation.
Hence (3/a) + (2/a) = 1
or, a = 5
So equation is: y + x = 5

Changing the line 2x - 3y + 12 = 0 to the intercept format of x/a + y/b = 1
2x + (-3)y = (-12)
or, (2x)/(-12) + {(-3)y}/(-12) = 1
or, x/(-6) + y/(4) = 1
So the intercepts on the x and y axes are (-6, 0) and (0, 4)

Since the points are on a straight line their slopes should be same.
Slope from 1st 2 points = (1 - 3)/(2 - a) = -2/(a - 2)
Slope from 2nd 2 points = (a - 1)/(5 - 2) = (a - 1)/3
Equating the two we get
-2/(a - 2) = (a - 1)/3
or, (a - 2)(a - 1) = -6
or, a² − 3a − 4 = 0
Solving for a we get a = 4 and -1

Coordinate where the line intercepts the x axis is (4, 0).
Slope of the line through (-2, 3) and (4, 0) is
m = (0 - 3)/(4 + 2) = -1/2
Equation of the line with m = -1/2 and passing through (-2, 3) is
(y - 3) = (-1/2)(x + 2)
or, 2y - 6 = -x - 2
or, x + 2y = 4

Given m = 1
Slope joining the points (2n, -3) and (5, n + 2) is
(n + 2 + 3)/(5 - 2n) = 1
or, n + 5 = 5 - 2n
or, n = 0