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.

$\hspace{3cm}m_1m_2=-1\\ \hspace{3.3cm}m_2=-\frac{1}{m_1}=\frac{-1}{-1/3}=3 \\ \text{The equation of line, from equation (i) is} \\ \hspace{3cm}(y-5)=3(x-0)\\ \hspace{4cm}y=3x+5$

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.

Given two lines $x+2y-5=0, x+3y-8=0$.