Question

part - a
find the equation of the line passing through the given points.

1. (1, -6) and (-7, 2)
2. (4, 5) and (-3, -1)
3. (-7, -9) and (-3, -4)
4. (2, -8) and (-6, -3)
5. (-9, 6) and (0, 3)
6. (3, 0) and (7, 1)
7. (8, 5) and (6, -7)
8. (2, 1) and (-4, -2)

part - b

1. find the equation of the line passing through the points (-5, -8) and (6, -4).
2. a line cuts the y - axis at (0, -6) and passes through the point (9, -3). find the equation of the line.

Explanation:

Let's solve the problem of finding the equation of the line passing through the points \((-5, -8)\) and \((6, -4)\) (from Part - B, Question 1) using the step - by - step format.

Step 1: Calculate the slope (\(m\))

The formula for the slope between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(m=\frac{y_2 - y_1}{x_2 - x_1}\).
Let \((x_1,y_1)=(-5,-8)\) and \((x_2,y_2)=(6,-4)\).
Then \(m=\frac{-4-(-8)}{6 - (-5)}=\frac{-4 + 8}{6 + 5}=\frac{4}{11}\).

Step 2: Use the point - slope form to find the equation of the line

The point - slope form of a line is \(y - y_1=m(x - x_1)\). We can use the point \((x_1,y_1)=(-5,-8)\) and \(m = \frac{4}{11}\).
Substitute these values into the point - slope formula:
\(y-(-8)=\frac{4}{11}(x - (-5))\)
\(y + 8=\frac{4}{11}(x + 5)\)

Step 3: Simplify the equation to slope - intercept form (\(y=mx + b\))

First, distribute \(\frac{4}{11}\) on the right - hand side:
\(y+8=\frac{4}{11}x+\frac{20}{11}\)
Then, subtract 8 from both sides. We know that \(8=\frac{88}{11}\), so:
\(y=\frac{4}{11}x+\frac{20}{11}-\frac{88}{11}\)
\(y=\frac{4}{11}x-\frac{68}{11}\)

Answer:

The equation of the line is \(y = \frac{4}{11}x-\frac{68}{11}\)