Determine the Linear Equation: Given Slope 1.5 and Point (3, 7.5)

Question

A line has a slope of 112 1\frac{1}{2} and passes through the point (3,712) (3,7\frac{1}{2}) .

Which expression corresponds to the line?

Video Solution

Solution Steps

00:00 Find the algebraic representation of the function
00:03 We'll use the straight line equation
00:09 We'll substitute the point according to the given data
00:15 We'll substitute the line's slope according to the given data
00:21 We'll continue solving to find the intersection point
00:35 We'll isolate intersection point B
00:41 This is the intersection point with the Y-axis
00:46 Now we'll substitute the intersection point and slope in the line equation
01:02 We'll factor out 1.5 from the parentheses
01:07 And this is the solution to the question

Step-by-Step Solution

To solve the problem of finding the equation of the line:

  • Step 1: Identify the given information: slope m=112=32 m = 1\frac{1}{2} = \frac{3}{2} , and point (x1,y1)=(3,712)=(3,152)(x_1, y_1) = (3, 7\frac{1}{2}) = (3, \frac{15}{2}).
  • Step 2: Use the point-slope formula yy1=m(xx1) y - y_1 = m(x - x_1) .
  • Step 3: Substitute the given slope and point into the formula: y152=32(x3) y - \frac{15}{2} = \frac{3}{2}(x - 3) .
  • Step 4: Distribute the slope on the right side: y152=32x92 y - \frac{15}{2} = \frac{3}{2}x - \frac{9}{2} .
  • Step 5: Add 152\frac{15}{2} to both sides to solve for y y : y=32x92+152 y = \frac{3}{2}x - \frac{9}{2} + \frac{15}{2} .
  • Step 6: Simplify the right side: y=32x+3 y = \frac{3}{2}x + 3 .
  • Step 7: Compare this expression to the provided choices to find a match. The form is the same as choice y=112(x+2) y=1\frac{1}{2}(x+2) , rewritten correctly as y=32(x+2) y = \frac{3}{2}(x + 2) .

Therefore, the expression that corresponds to the line is y=112(x+2) y = 1\frac{1}{2}(x+2) .

Answer

y=112(x+2) y=1\frac{1}{2}(x+2)