Line L1 has equation 2x -3y+12=0

Gradient of L1 is 2/3x

The line L1 crosses the x axis at the point A and the y axis at the point B.

The line L2 is perpendicular to L1 and passes through B.

A) find and equation of L2

The line L2 crosses the x axis at the point C

B) find the area of triangle ABC

Thankyou

A) Well, since L2 is perpendicular to L1, we know that the product of their gradients is -1. So, the gradient of L2 will be the negative reciprocal of the gradient of L1, which is -3/2x.

To find the equation of L2, we need to know a point it passes through. We know that L2 passes through point B, which is where L1 intersects the y-axis. The coordinates of point B are (0, b), where b is the y-coordinate when x = 0 on L1.

We can substitute the coordinates of B into the equation of L2 to find the value of b, which is the y-intercept of L2.

So the equation of L2 would be y = -3/2x + b, where b is the value we find when we substitute the coordinates of B.

B) Now, to find the area of triangle ABC, we need to find the coordinates of points A and C.

Point A is where L1 intersects the x-axis, so its y-coordinate would be 0. We can find the x-coordinate by substituting y = 0 into the equation of L1 and solving for x.

Point C is where L2 intersects the x-axis, so its y-coordinate would also be 0. Similarly, we can find the x-coordinate by substituting y = 0 into the equation of L2 and solving for x.

Once we have the coordinates of points A, B, and C, we can use the formula for the area of a triangle to find the area of triangle ABC.

Remember, the formula for the area of a triangle is 0.5 * base * height. In this case, the base would be the distance between points A and C (which is the horizontal distance), and the height would be the perpendicular distance from point B to line AC.

I hope that helps and puts a smile on your face!

To find the equation of line L2, we need to find its gradient. Since L2 is perpendicular to L1, we can use the negative reciprocal of the gradient of L1, which is -3/2.

A) Equation of line L2: y - y1 = m(x - x1)

Since L2 passes through point B, which is the y-intercept of L1, its coordinates are (0, -4).

Using the coordinates (0, -4) and the gradient -3/2, the equation of line L2 becomes:

y - (-4) = (-3/2)(x - 0)
y + 4 = (-3/2)x

Simplifying, we get the equation of line L2 as:
2y + 8 = -3x
3x + 2y + 8 = 0

B) To find the area of triangle ABC, we need to determine the coordinates of points A and C.

Since line L1 crosses the x-axis at point A, its y-coordinate is 0. To find the x-coordinate, substitute y = 0 into the equation of L1:
2x - 3(0) + 12 = 0
2x + 12 = 0
2x = -12
x = -6

Therefore, point A has the coordinates (-6, 0).

To find the point C, we know that line L2 crosses the x-axis at that point. So, its y-coordinate is 0. To find the x-coordinate, substitute y = 0 into the equation of L2:
3x + 2(0) + 8 = 0
3x + 8 = 0
3x = -8
x = -8/3

Therefore, point C has the coordinates (-8/3, 0).

The area of triangle ABC can be calculated using the Shoelace Formula:

Area = 1/2 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Substituting the coordinates of points A, B, and C, the area becomes:

Area = 1/2 * |-6(0 - 0) + 0(-4 - 0) + (-8/3)(0 - (-4))|
Area = 1/2 * |-8/3|
Area = 4/3

Therefore, the area of triangle ABC is 4/3 square units.

A) To find the equation of line L2, we need to determine its gradient first. Since L2 is perpendicular to L1, the product of their gradients must be -1.

Given the gradient of L1 as 2/3x, we can find the gradient of L2 by taking the negative reciprocal: -(3/2x).

Now, we know the gradient of L2, but we need a point on L2 to find its equation. We are given that L2 passes through point B, which is the y-intercept of L1. We know that the y-intercept occurs when x=0, so substituting x=0 into the equation of L1 gives us:

2(0) - 3y + 12 = 0
-3y + 12 = 0
-3y = -12
y = 4

Therefore, point B is (0, 4). We now have the gradient and a point on L2, so we can use the point-slope form of the equation for a straight line to find the equation of L2.

y - y1 = m(x - x1)

Plugging in the values, we get:

y - 4 = -(3/2)(x - 0)

Simplifying, we have:

y - 4 = -(3/2)x

This is the equation of line L2.

B) To find the area of triangle ABC, we need to know the coordinates of points A and C.

Since line L1 crosses the x-axis at point A, we know that y=0 when L1 intersects the x-axis. Plugging in y=0 into the equation of L1 gives us:

2x - 3(0) + 12 = 0
2x + 12 = 0
2x = -12
x = -6

Therefore, point A is (-6, 0).

To find point C, we need to determine where line L2 crosses the x-axis. We already have the equation for L2:

y - 4 = -(3/2)x

Substituting y=0 into the equation gives us:

0 - 4 = -(3/2)x
-4 = -(3/2)x
4 = (3/2)x
8/3 = x

Therefore, point C is (8/3, 0).

Now that we have the coordinates of points A, B, and C, we can calculate the area of triangle ABC.

The area of a triangle can be found using the formula:

Area = (1/2) * base * height

In this case, the base of the triangle is the distance between point A and point C, which can be found using the distance formula:

Distance = √[(x2 - x1)^2 + (y2 - y1)^2]

Plugging in the values, we have:

Distance = √[(-6 - 8/3)^2 + (0 - 0)^2]
Distance = √[( -26/3 )^2]
Distance = 26/3 units

The height of the triangle is the y-coordinate of point B, which is 4 units.

Now, we can calculate the area using the formula:

Area = (1/2) * base * height
Area = (1/2) * (26/3) * 4
Area = (26/3) * 2
Area = 52/3 square units

Therefore, the area of triangle ABC is 52/3 square units.

L1 has intercepts at A:(-6,0) and B:(0,4)

L2 has gradient -3/2 and passes through (0,4), so L2 is

y-4 = (-3/2)(x-0)
3x+2y-8 = 0
L2 passes through B:(0,4) and C:(8/3,0)

So, if you plot the points, you can see that you have a triangle with base along the x-axis, and its third point on the y-axis.