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To solve for the equation of the straight line passing through the point of intersection of the lines [tex]\( y = 2x + 3 \)[/tex] and [tex]\( x + 2y = -4 \)[/tex], and is perpendicular to the line passing through the points [tex]\((2, 4)\)[/tex] and [tex]\((-1, 5)\)[/tex], follow the steps below:
### Step 1: Finding the Intersection Point
We need to find the intersection point (x, y) of the lines [tex]\( y = 2x + 3 \)[/tex] and [tex]\( x + 2y = -4 \)[/tex].
By solving these equations simultaneously:
1. [tex]\( y = 2x + 3 \)[/tex]
2. [tex]\( x + 2y = -4 \)[/tex]
Substitute [tex]\( y = 2x + 3 \)[/tex] into the second equation:
[tex]\[ x + 2(2x + 3) = -4 \][/tex]
[tex]\[ x + 4x + 6 = -4 \][/tex]
[tex]\[ 5x + 6 = -4 \][/tex]
[tex]\[ 5x = -10 \][/tex]
[tex]\[ x = -2 \][/tex]
Now, substitute [tex]\( x = -2 \)[/tex] back into [tex]\( y = 2x + 3 \)[/tex]:
[tex]\[ y = 2(-2) + 3 \][/tex]
[tex]\[ y = -4 + 3 \][/tex]
[tex]\[ y = -1 \][/tex]
So, the intersection point is [tex]\((-2, -1)\)[/tex].
### Step 2: Finding the Slope of the Line Through [tex]\((2, 4)\)[/tex] and [tex]\((-1, 5)\)[/tex]
Calculate the slope [tex]\( m \)[/tex] of the line passing through the points [tex]\((2, 4)\)[/tex] and [tex]\((-1, 5)\)[/tex]:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
[tex]\[ m = \frac{5 - 4}{-1 - 2} \][/tex]
[tex]\[ m = \frac{1}{-3} \][/tex]
[tex]\[ m = -\frac{1}{3} \][/tex]
### Step 3: Finding the Perpendicular Slope
The slope of a line perpendicular to a given line is the negative reciprocal of the slope of that line. Thus, the perpendicular slope is:
[tex]\[ m_{\perp} = -\frac{1}{-\frac{1}{3}} \][/tex]
[tex]\[ m_{\perp} = 3 \][/tex]
### Step 4: Equation of the Perpendicular Line
Using the point-slope form of a line's equation, [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\((x_1, y_1) = (-2, -1)\)[/tex] and [tex]\( m = 3 \)[/tex]:
[tex]\[ y - (-1) = 3(x - (-2)) \][/tex]
[tex]\[ y + 1 = 3(x + 2) \][/tex]
[tex]\[ y + 1 = 3x + 6 \][/tex]
[tex]\[ y = 3x + 6 - 1 \][/tex]
[tex]\[ y = 3x + 5 \][/tex]
Therefore, the equation of the line passing through the point of intersection and perpendicular to the line passing through the points [tex]\((2, 4)\)[/tex] and [tex]\((-1, 5)\)[/tex] is:
[tex]\[ \boxed{y = 3x + 5} \][/tex]
### Step 1: Finding the Intersection Point
We need to find the intersection point (x, y) of the lines [tex]\( y = 2x + 3 \)[/tex] and [tex]\( x + 2y = -4 \)[/tex].
By solving these equations simultaneously:
1. [tex]\( y = 2x + 3 \)[/tex]
2. [tex]\( x + 2y = -4 \)[/tex]
Substitute [tex]\( y = 2x + 3 \)[/tex] into the second equation:
[tex]\[ x + 2(2x + 3) = -4 \][/tex]
[tex]\[ x + 4x + 6 = -4 \][/tex]
[tex]\[ 5x + 6 = -4 \][/tex]
[tex]\[ 5x = -10 \][/tex]
[tex]\[ x = -2 \][/tex]
Now, substitute [tex]\( x = -2 \)[/tex] back into [tex]\( y = 2x + 3 \)[/tex]:
[tex]\[ y = 2(-2) + 3 \][/tex]
[tex]\[ y = -4 + 3 \][/tex]
[tex]\[ y = -1 \][/tex]
So, the intersection point is [tex]\((-2, -1)\)[/tex].
### Step 2: Finding the Slope of the Line Through [tex]\((2, 4)\)[/tex] and [tex]\((-1, 5)\)[/tex]
Calculate the slope [tex]\( m \)[/tex] of the line passing through the points [tex]\((2, 4)\)[/tex] and [tex]\((-1, 5)\)[/tex]:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
[tex]\[ m = \frac{5 - 4}{-1 - 2} \][/tex]
[tex]\[ m = \frac{1}{-3} \][/tex]
[tex]\[ m = -\frac{1}{3} \][/tex]
### Step 3: Finding the Perpendicular Slope
The slope of a line perpendicular to a given line is the negative reciprocal of the slope of that line. Thus, the perpendicular slope is:
[tex]\[ m_{\perp} = -\frac{1}{-\frac{1}{3}} \][/tex]
[tex]\[ m_{\perp} = 3 \][/tex]
### Step 4: Equation of the Perpendicular Line
Using the point-slope form of a line's equation, [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\((x_1, y_1) = (-2, -1)\)[/tex] and [tex]\( m = 3 \)[/tex]:
[tex]\[ y - (-1) = 3(x - (-2)) \][/tex]
[tex]\[ y + 1 = 3(x + 2) \][/tex]
[tex]\[ y + 1 = 3x + 6 \][/tex]
[tex]\[ y = 3x + 6 - 1 \][/tex]
[tex]\[ y = 3x + 5 \][/tex]
Therefore, the equation of the line passing through the point of intersection and perpendicular to the line passing through the points [tex]\((2, 4)\)[/tex] and [tex]\((-1, 5)\)[/tex] is:
[tex]\[ \boxed{y = 3x + 5} \][/tex]
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