Problem 106 - Algebra and Trigonometry

Question

Write an equation in slope-intercept form for the line passing through \((1,-1)\) and perpendicular to the line whose equation is \(x+10 y-13=0\)

Step-by-Step Solution

Answer

The equation of the line in slope-intercept form is \(y = 10x - 11\).

Step 1: Find the slope of the given line

The equation of the given line is \(x + 10y - 13 = 0\), from which we can derive the slope by rearranging the equation in the slope-intercept form \(y = mx + b\). So, the original equation can be rewritten as \(y = -\frac{1}{10}x + \frac{13}{10}\). The slope ('m') of this line is -1/10.

Step 2: Find the slope of the required line

We know that the slopes of two perpendicular lines are negative reciprocals of each other. So, the slope of the line we're looking for is the negative reciprocal of -1/10, which is 10.

Step 3: Use the point-slope form to write the equation of the required line

We can use the point-slope form, which is \(y - y1 = m(x - x1)\), where \((x1, y1)\) is the point the line passes through, and 'm' is the slope of the line. Substituting the given point \((1,-1)\) and slope '10', we get: \(y - (-1) = 10(x - 1)\).

Step 4: Write the equation in slope-intercept form

Finally, we simplify the equation from the previous step to obtain it in slope-intercept form. Simplifying, we get \(y = 10x - 11\).