What is the distance from the point (-1, -3) to the line y = 2?
How do i do this?
In ABC, what is the angle opposite ? I also need help with this one.
In general, the distance of the point
P0 : (x0,y0)
from the line
L : ax+by+c = 0
is given by:
Distance = |ax0+by0+c|/sqrt(a²+b²) ...(1)
where the pair of | | in the numerator represents the absolute value.
In the given case,
P0 : (-1,-3)
or x0=-1, y0=-3
and
L : y - 2 = 0
or a=0, b=1, c=-2
Substitute these values in (1) to get the distance required.
To find the distance from a point to a line, you can use the formula for the distance between a point and a line. In this case, we have the point (-1, -3) and the line y = 2.
Step 1: Rewrite the equation of the line in the slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept. In this case, since the line is horizontal (y = 2), the slope is 0 and the y-intercept is 2.
Step 2: Use the formula for the distance between a point (x1, y1) and a line (Ax + By + C = 0):
distance = |Ax1 + By1 + C| / sqrt(A^2 + B^2)
In this case, the equation of the line is y = 2, which can be rewritten as 0x + 1y - 2 = 0. So, A = 0, B = 1, and C = -2.
Step 3: Substitute the values into the formula and calculate the distance:
distance = |0(-1) + 1(-3) - 2| / sqrt(0^2 + 1^2)
distance = |-3 - 2| / sqrt(0 + 1)
distance = |-5| / sqrt(1)
distance = 5 / 1
distance = 5
Therefore, the distance from the point (-1, -3) to the line y = 2 is 5 units.