The straight line L1 passes through the points with coordinates (6, 5) and (10, 3) The straight line L2 passes through the origin and has gradient -3
The lines L1 and L2 intersect at point P
Find the coordinates of P.

The coordinates of point p are:

x = -3.2

y = 9.6

Step-by-step explanation:

A linear relationship can be written as:

y = a*x + b

where a is the slope and b is the y-axis intercept.

For a line that passes through the points (x1, y1) and (x2, y2), the slope can be written as:

a = (y2 - y1)/(x2 - x1).

Then if we know that L1 passes through the points (6, 5) and (10, 3) the slope of this line will be:

a = (3 - 5)/(10 - 6) = -0.5

Then this line will be something like:

y = -0.5*x + b

To find the value of b we can just replace the values of one of the points in the above equation. For example if we use the point (6, 5), this means that:

x = 6

y = 5

Then we get:

5 = -0.5*6 + b

5 = -3 + b

5 + 3 = b

8 = b

Then the equation for line L1 is:

y = -0.5*x + 8

Now, for line L2 we know that it has a gradient -3 (for linear equations, the gradient is the same as the slope)

Then line L2 has a slope equal to -3

y = -3*x + b

And we know that this line passes through the origin, (0, 0)

Then if we replace the values of that point in our equation, we get:

0 = -3*0 + b

0 = b

Then line L2 is:

y = -3*x

Then our two lines are:

y = -0.5*x + 8

y = -3*x

Now we want to find the point where these lines intersect. The lines will intersect in one point that belongs to both lines, then we can use the relation:

-0.5*x + 8 = y = -3*x

-0.5*x + 8 = -3*x

now we can solve this for x.

-0.5*x + 8 = -3*x

8 = -3*x + 0.5*x

8 = -2.5*x

8/-2.5 = x = -3.2

Now we need to evaluate one of the lines on this value, i will use line L2

y = -3*(-3.2) = 9.6

Then we can conclude that the lines do intersect in the point:

x = -3.2

y = 9.6

this point is written as: (-3.2, 9.6)

x = 25

step-by-step explanation:

a^2+b^2=c^2

7^2+24^2=c^2

49+576=625

sqrt625=25

x=25