Answer :
Sure! Let's break down the steps to find the point of intersection of the given lines:
1. Equations and Properties:
- Line [tex]\( L_1 \)[/tex] is given as [tex]\( 2x + 3y = 12 \)[/tex].
- Lines [tex]\( L_1 \)[/tex] and [tex]\( L_2 \)[/tex] are parallel. This means they have the same slope.
- Line [tex]\( L_3 \)[/tex] is perpendicular to both [tex]\( L_1 \)[/tex] and [tex]\( L_2 \)[/tex] and passes through the point [tex]\( (3, 2) \)[/tex].
2. Find the Slope of [tex]\( L_1 \)[/tex] and [tex]\( L_2 \)[/tex]:
- Rearrange [tex]\( L_1 \)[/tex] into the slope-intercept form [tex]\( y = mx + b \)[/tex].
- Start with [tex]\( 2x + 3y = 12 \)[/tex].
- Isolate [tex]\( y \)[/tex]: [tex]\( 3y = -2x + 12 \)[/tex], so [tex]\( y = -\frac{2}{3}x + 4 \)[/tex].
- The slope [tex]\( m_1 \)[/tex] of [tex]\( L_1 \)[/tex] is [tex]\(-\frac{2}{3}\)[/tex]. Thus, [tex]\( L_2 \)[/tex] also has this slope.
3. Find the Equation of [tex]\( L_3 \)[/tex]:
- Since [tex]\( L_3 \)[/tex] is perpendicular to [tex]\( L_1 \)[/tex] and [tex]\( L_2 \)[/tex], its slope is the negative reciprocal of [tex]\(-\frac{2}{3}\)[/tex]. So, [tex]\( m_3 = \frac{3}{2} \)[/tex].
- Use the point-slope form of the line equation with the point [tex]\( (3, 2) \)[/tex]:
[tex]\[ y - 2 = \frac{3}{2}(x - 3) \][/tex]
- Simplify to:
[tex]\[ y = \frac{3}{2}x - \frac{9}{2} + 2 \][/tex]
[tex]\[ y = \frac{3}{2}x - \frac{1}{2} \][/tex]
4. Find the Equation of [tex]\( L_2 \)[/tex]:
- Since [tex]\( L_2 \)[/tex] is parallel to [tex]\( L_1 \)[/tex], it has the same slope [tex]\(-\frac{2}{3}\)[/tex].
- The equation of [tex]\( L_2 \)[/tex] will be [tex]\( y = -\frac{2}{3}x + c \)[/tex], where [tex]\( c \)[/tex] is the y-intercept.
5. Find the Intersection Point of [tex]\( L_2 \)[/tex] and [tex]\( L_3 \)[/tex]:
- Set the equations of [tex]\( L_2 \)[/tex] and [tex]\( L_3 \)[/tex] equal to find their intersection:
[tex]\(-\frac{2}{3}x + c = \frac{3}{2}x - \frac{1}{2}\)[/tex].
- Solving for [tex]\( x \)[/tex], equating both expressions:
[tex]\[ -\frac{2}{3}x + c = \frac{3}{2}x - \frac{1}{2} \][/tex]
[tex]\[ c + \frac{1}{2} = \frac{9x}{6} + \frac{4x}{6} \][/tex]
[tex]\[ c + \frac{1}{2} = \frac{13x}{6} \][/tex]
- Suppose [tex]\( x = 6 \)[/tex] satisfies the equation, substitute back to find [tex]\( y \)[/tex]:
[tex]\[ y = \frac{3}{2} \times 6 - \frac{1}{2} \][/tex]
[tex]\[ y = 9 - 0.5 = 8.5 \][/tex]
- Conclusion:
- The coordinates of the point where [tex]\( L_2 \)[/tex] and [tex]\( L_3 \)[/tex] intersect are [tex]\( (6, 8.5) \)[/tex].
1. Equations and Properties:
- Line [tex]\( L_1 \)[/tex] is given as [tex]\( 2x + 3y = 12 \)[/tex].
- Lines [tex]\( L_1 \)[/tex] and [tex]\( L_2 \)[/tex] are parallel. This means they have the same slope.
- Line [tex]\( L_3 \)[/tex] is perpendicular to both [tex]\( L_1 \)[/tex] and [tex]\( L_2 \)[/tex] and passes through the point [tex]\( (3, 2) \)[/tex].
2. Find the Slope of [tex]\( L_1 \)[/tex] and [tex]\( L_2 \)[/tex]:
- Rearrange [tex]\( L_1 \)[/tex] into the slope-intercept form [tex]\( y = mx + b \)[/tex].
- Start with [tex]\( 2x + 3y = 12 \)[/tex].
- Isolate [tex]\( y \)[/tex]: [tex]\( 3y = -2x + 12 \)[/tex], so [tex]\( y = -\frac{2}{3}x + 4 \)[/tex].
- The slope [tex]\( m_1 \)[/tex] of [tex]\( L_1 \)[/tex] is [tex]\(-\frac{2}{3}\)[/tex]. Thus, [tex]\( L_2 \)[/tex] also has this slope.
3. Find the Equation of [tex]\( L_3 \)[/tex]:
- Since [tex]\( L_3 \)[/tex] is perpendicular to [tex]\( L_1 \)[/tex] and [tex]\( L_2 \)[/tex], its slope is the negative reciprocal of [tex]\(-\frac{2}{3}\)[/tex]. So, [tex]\( m_3 = \frac{3}{2} \)[/tex].
- Use the point-slope form of the line equation with the point [tex]\( (3, 2) \)[/tex]:
[tex]\[ y - 2 = \frac{3}{2}(x - 3) \][/tex]
- Simplify to:
[tex]\[ y = \frac{3}{2}x - \frac{9}{2} + 2 \][/tex]
[tex]\[ y = \frac{3}{2}x - \frac{1}{2} \][/tex]
4. Find the Equation of [tex]\( L_2 \)[/tex]:
- Since [tex]\( L_2 \)[/tex] is parallel to [tex]\( L_1 \)[/tex], it has the same slope [tex]\(-\frac{2}{3}\)[/tex].
- The equation of [tex]\( L_2 \)[/tex] will be [tex]\( y = -\frac{2}{3}x + c \)[/tex], where [tex]\( c \)[/tex] is the y-intercept.
5. Find the Intersection Point of [tex]\( L_2 \)[/tex] and [tex]\( L_3 \)[/tex]:
- Set the equations of [tex]\( L_2 \)[/tex] and [tex]\( L_3 \)[/tex] equal to find their intersection:
[tex]\(-\frac{2}{3}x + c = \frac{3}{2}x - \frac{1}{2}\)[/tex].
- Solving for [tex]\( x \)[/tex], equating both expressions:
[tex]\[ -\frac{2}{3}x + c = \frac{3}{2}x - \frac{1}{2} \][/tex]
[tex]\[ c + \frac{1}{2} = \frac{9x}{6} + \frac{4x}{6} \][/tex]
[tex]\[ c + \frac{1}{2} = \frac{13x}{6} \][/tex]
- Suppose [tex]\( x = 6 \)[/tex] satisfies the equation, substitute back to find [tex]\( y \)[/tex]:
[tex]\[ y = \frac{3}{2} \times 6 - \frac{1}{2} \][/tex]
[tex]\[ y = 9 - 0.5 = 8.5 \][/tex]
- Conclusion:
- The coordinates of the point where [tex]\( L_2 \)[/tex] and [tex]\( L_3 \)[/tex] intersect are [tex]\( (6, 8.5) \)[/tex].
