Absolutely, let’s delve into solving the problems step by step:
### Problem 74
The question requires us to find the length of the perpendicular from the point \((0,0)\) to the line \(3x + 4y - 10 = 0\).
We can use the formula for the distance \(d\) from a point \((x_0, y_0)\) to a line \(ax + by + c = 0\), which is given by:
[tex]\[
d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}
\][/tex]
1. Substitute \((x_0, y_0) = (0,0)\) into the formula:
[tex]\[
d = \frac{|3 \cdot 0 + 4 \cdot 0 - 10|}{\sqrt{3^2 + 4^2}}
\][/tex]
2. Simplify the expression in the numerator:
[tex]\[
d = \frac{|0 + 0 - 10|}{\sqrt{9 + 16}} = \frac{|-10|}{\sqrt{25}} = \frac{10}{5} = 2
\][/tex]
So, the length of the perpendicular from \((0,0)\) to the line \(3x + 4y - 10 = 0\) is \(2\).
The correct answer is \(\boxed{2}\).
### Problem 75
We need to find the value of \(x\) for which the matrix
[tex]\[
A = \begin{pmatrix}
6 & x-2 \\
3 & x
\end{pmatrix}
\][/tex]
has no inverse.
A matrix has no inverse if its determinant is zero. The determinant of matrix \(A\) can be found by:
[tex]\[
\text{det}(A) = 6 \cdot x - 3 \cdot (x - 2)
\][/tex]
1. Calculate the determinant:
[tex]\[
\text{det}(A) = 6x - 3(x - 2)
\][/tex]
Simplify the expression:
[tex]\[
\text{det}(A) = 6x - 3x + 6 = 3x + 6
\][/tex]
2. Set the determinant equal to zero:
[tex]\[
3x + 6 = 0
\][/tex]
3. Solve for \(x\):
[tex]\[
3x = -6 \implies x = -2
\][/tex]
So, the value of \(x\) for which the matrix has no inverse is \( -2 \).
The correct answer is \(\boxed{-2}\).
Each problem has been carefully broken down and solved with clear steps to arrive at the correct answers.
