Q.$P(a, b)$ is the mid-point of a line segment between axes. Show that equation of the line is $\dfrac{x}{a} + \dfrac{y}{b} = 2$.

The equation of the $x$-axis is $y=0$.

The $x$-axis is the set of all points where the $y$-coordinate is zero, regardless of the value of $x$ — i.e., points of the form $(x, 0)$.

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The $y$-intercept is $-4$.

The $y$-intercept of a line is the value of $y$ when $x=0$.

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$y=0,\ z=0$ — option (d).

The $x$-axis is the set of points where both the $y$-coordinate and $z$-coordinate are zero (only $x$ varies). So its equations i …

A line parallel to the $x$-axis has a constant $y$-coordinate equal to the $y$-coordinate of the given point.

Any line parallel to the $x$-axis has the form $y = k$ for some constant $k$.

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Since every point on the x-axis has y-coordinate 0, its equation is $y=0$.

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The line $3x+2y=6$ cuts the axes at $x=2$ and $y=3$.

$x$-intercept (set $y=0$): $3x=6\Rightarrow x=2$. …

The required line is $2x-3y+6=0$, matching option (b).

A line with $x$-intercept $a=-3$ and $y$-intercept $b=2$ has intercept form (NCERT Class 11 Straight Lines):

$\dfrac{x}{-3}+\dfrac{y}{2}=1.$

Multiply through by $6$: $-2x+3y=6$, i.e. $2x-3y+6=0$. Check: at $y=0,\ x=-3$; at $x=0,\ y=2$.

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Setting $x=0$ in the intercept form directly gives $y = 4$.

The line is given in intercept form:

$\frac{x}{3} + \frac{y}{4} = 1$

A point on the $y$-axis has $x = 0$. Substituting: …

The line is $2x-3y+6=0$.

With x-intercept $-3$ and y-intercept $2$, the intercept form is $\dfrac{x}{-3}+\dfrac{y}{2}=1$.

Multiply by $6$: $-2x+3y=6$, i.e. $2x-3y+6=0$.

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Setting y=0 in $4x+3y=12$ gives the point $(3,0)$.

A line meets the x-axis where $y=0$.

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The line $3x+2y=6$ cuts the axes at $x=2$ and $y=3$.

For the $x$-intercept, set $y=0$: $3x=6 \Rightarrow x=2$. …