Coordinate Geometry - Coordinates as Perpendicular Distances-advanced

Find the coordinates of a point $P$ which lies in the third quadrant, such that its perpendicular distance from the $x$-axis is $5$ units and its perpendicular distance from the $y$-axis is $3$ units.

1. In the third quadrant, both the $x$-coordinate and $y$-coordinate are negative.
2. The perpendicular distance from the $x$-axis is the absolute value of the $y$-coordinate: $|y| = 5$. Since it is in the third quadrant, $y = -5$.
3. The perpendicular distance from the $y$-axis is the absolute value of the $x$-coordinate: $|x| = 3$. Since it is in the third quadrant, $x = -3$.
4. Therefore, the coordinates are $(-3, -5)$.

The distance from an axis determines the magnitude of the coordinate belonging to the other axis. The quadrant determines the sign.

A point $M$ is at a distance of $7$ units from the $y$-axis and lies on the $x$-axis to the left of the origin. Find its coordinates.

1. Since the point lies on the $x$-axis, its $y$-coordinate must be $0$.
2. The distance from the $y$-axis is $7$ units, which means $|x| = 7$.
3. Since the point is to the left of the origin, the $x$-coordinate must be negative, so $x = -7$.
4. The coordinates are $(-7, 0)$.

Points on the $x$-axis always have an ordinate of $0$. Distance from the $y$-axis gives the abscissa.

If the perpendicular distance of a point $Q(a, b)$ from the $x$-axis is $4$ units and from the $y$-axis is $6$ units, and the point lies in the second quadrant, find the value of $2a + b$.

1. In the second quadrant, $x < 0$ and $y > 0$.
2. Distance from $x$-axis $= |b| = 4$. Since $y > 0$, $b = 4$.
3. Distance from $y$-axis $= |a| = 6$. Since $x < 0$, $a = -6$.
4. We need to find $2a + b$: $2(-6) + 4$ $-12 + 4$ $-8$

The coordinates are found using the quadrant signs and distances, then substituted into the linear expression.

1. In the fourth quadrant, the $x$-coordinate is positive and the $y$-coordinate is negative.
2. Perpendicular distance from the $y$-axis is $|x| = 4$. Since $x > 0$, $x = 4$.
3. Perpendicular distance from the $x$-axis is $|y| = 2$. Since $y < 0$, $y = -2$.
4. The coordinates of point $R$ are $(4, -2)$.
5. Calculate $x^2 - y$: $4^2 - (-2) = 16 + 2 = 18$.

We use the quadrant signs (Quadrant IV: $+, -$) and the definition of coordinates as distances from the axes to locate the point.

1. Distance from the $y$-axis is $0$, so the $x$-coordinate is $0$. The point lies on the $y$-axis.
2. Distance from the $x$-axis is $6$. Since it lies on the positive $y$-axis, $y = 6$.
3. Coordinates of $S$ are $(0, 6)$.
4. Distance from $T(0, -2)$ to $S(0, 6)$ along the $y$-axis is $|6 - (-2)| = |6 + 2| = 8$ units.

When a point is on an axis, one of its perpendicular distances is zero. Here, the point lies on the vertical axis, and we calculate the linear distance between two points on the same axis.