Straight Lines and Coordinates

This is a coordinate geometry problem requiring finding intersection point location. Solving the system: 3x - 5y = 10 and 2x - 10y = -3. From the second equation: x = 5y - 1.5. Substituting into first: 3(5y - 1.5) - 5y = 10, giving 15y - 4.5 - 5y = 10, so 10y = 14.5, and y = 1.45. Then x = 5(1.45) - 1.5 = 7.25 - 1.5 = 5.75. The intersection point (5.75, 1.45) has both coordinates positive, placing it in the first quadrant.

Since x = -2, substitute this into the first equation: 2(-2) + 3y = 5. This simplifies to -4 + 3y = 5, then 3y = 9, so y = 3. The intersection point is (-2, 3).

L1 has slope (3-2)/(2-0) = 1/2. Since L1 and L2 are perpendicular, L2's slope must be -2 (negative reciprocal). Using point-slope form through (2,3): y-3 = -2(x-2). Checking point (3,1): 1-3 = -2 and -2(3-2) = -2, so it satisfies the equation. The other options do not satisfy this line equation.

The slope of ax+by=c is -a/b. The slope of bx-ay=c is b/a. The product of slopes is (-a/b) × (b/a) = -1, which means the lines are perpendicular to each other.

Parallel lines have the same coefficients for x and y. The given line 3x + 7y = 10 has slope -3/7. Substituting point (4, 8) into 3x + 7y = c gives c = 3(4) + 7(8) = 12 + 56 = 68. The parallel line through (4, 8) is 3x + 7y = 68.

Slope = difference in y coordinates/difference in x coordinates = 3

k = 15/4 is the right option.

The two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 will be parallel if $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. ^_$\because$        $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$ and $\frac{c_1}{c_2} = \frac{5}{12} $. ^{_$\therefore$}        $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ Hence, 2x + 3y = 5 and 4x + 6y = 12 are parallel lines.

For the pair of equations to represent coincident lines, $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ In lines x + 2y = 7 and 2x + 4y = 14, $\frac{a_1}{a_2} = \frac{1}{2}, \frac{b_1}{b_2} =\frac{2}{4} =\frac{1}{2}, \frac{c_1}{c_2} = \frac{7}{14} = \frac{1}{2}$ So, _{$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$}

Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel when $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. In lines x – y = 7 and 2x – 2y = 15, $\frac{a_1}{a_2} = \frac{1}{2}, \frac{b_1}{b_2} =\frac{-1}{-2} , \frac{c_1}{c_2} = \frac{7}{15}$ ^{_$\therefore$}        $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

A pair of lines is coincident, if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. In lines, x + 2y = 3 and 4x + 8y = 12 $\frac{a_1}{a_2} = \frac{1}{4}, \frac{b_1}{b_2} =\frac{2}{8} = \frac{1}{4} , \frac{c_1}{c_2} = \frac{3}{12} = \frac{1}{4}$ $\therefore$   $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

For two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 to be parallel, $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. In lines x – y = 17 and 2x – 2y = 38, $\frac{a_1}{a_2} = \frac{1}{2}, \frac{b_1}{b_2} =\frac{-1}{-2} = \frac{1}{2} , \frac{c_1}{c_2} = \frac{3}{12} = \frac{17}{38}$ Therefore, ^{_{$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$}} Hence, the lines are parallel.

Let the given point  be $A=(−5,4)$ and the given lines be $l _1 \rightarrow x+y+1=0$ and  $l _2\rightarrow x+y-1=0$

The  point $A$ lies on $l _1$

If segment $AM\perp l _2$ and $M$ lies on $l _2$, then, the distance $ AM$ is given by,

$\Rightarrow AM=\dfrac{|−5+4−1|}{\sqrt{1^2+1^2}}=\dfrac{2}{\sqrt 2}=\sqrt 2$

$\Rightarrow $ This means that if $B$ is any point  on $l _2$ then  $AB>AM$. No line other than $AM$ cuts off an intercept of

length $\sqrt 2$ between $l _1$ and $l _2$.

$\Rightarrow $To determine the equation of $AM$, we need to find the co-ordinates of the Point $M$

Since, $AM\perp l _2$ and  the slope $l _2$ is $−1$, the slope of$AM$ must be $1$.  Also  $A(−5,4)$ lies on $AM$

By the point slope formula, the equation of the required line is

$\Rightarrow  y−4=1(x−(−5))$

$\Rightarrow y-4=x+5$

$\Rightarrow x−y+9=0$

The slopes of the given lines are m₁ = 3/4 and m₂ = 12/5. For a line to be equally inclined to both, its slope m must satisfy |(m-m₁)/(1+mm₁)| = |(m-m₂)/(1+mm₂)|. Solving gives two possible slopes: m = -7/9 (internal bisector) or m = 9/7 (external bisector). Using point (4,5) with slope -7/9: y-5 = (-7/9)(x-4), giving 9y-45 = -7x+28, or 7x+9y=73.

Let the given point be $A=(-5,4)$ and the given lines be,