Answer:
6 units
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The length of the other altitude 6 units. Then the correct option is B.
Let (x₁, y₁) be the point and ax + by + c = 0 be the line.
Then the distance between a point and a line will be
[tex]\rm d = \dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}[/tex]
The equation of the line will be
(y – 6) = [(6 – 2)/(10 – 7)](x – 10)
y – 6 = 4/3(x – 10)
3y – 18 = 4x – 40
4x – 3y – 22 = 0
Then the distance between point (4, 8) and line 4x – 3y – 22 = 0 will be
[tex]\rm d = \dfrac{|4*4 - 3*8 - 22|}{\sqrt{4^2 + (-3)^2}}\\\\\\d = \dfrac{|4*4 - 3*8 - 22|}{\sqrt{4^2 + (-3)^2}}\\\\\\d = \dfrac{|16 - 24 - 22|}{\sqrt{16+9}}\\\\\\d = \dfrac{|-30|}{\sqrt{25}}\\[/tex]
On simplifying, we have
d = 30 / 5
d = 6 units
Then the correct option is B.
Learn more about the distance between a point and a line here;
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